# solve Sinx=Cos2x-1 for 0¡Üx<2¦Ð

24,954 results-
## Trig--check answer

Solve the equation of the interval (0, 2pi) cosx=sinx I squared both sides to get :cos²x=sin²x Then using tri indentites I came up with cos²x=1-cos²x Ended up with 2cos²x=1 Would the answer be cos²x=1/2??? -
## Math Help

Hello! Can someone please check and see if I did this right? Thanks! :) Directions: Find the exact solutions of the equation in the interval [0,2pi] cos2x+sinx=0 My answer: cos2x+sinx=cos^2x-sin^2x+sinx =1-sin^2x-sin^2x+sinx =-2sin^2x+sinx+1=0 -
## math

the problem is 2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is what i've done so far: 2cos^2x+sinx-1=0 2cos^2x-1+sinx=0 cos2x + sinx =0 1 - 2sin^2x + sinx = 0 -2sin^2x+sinx-1=0 -
## math

cos2x-sinx=0. Factor and solve for sinx -
## Precalculus

Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π). 8. cos2x=cosx 10. 2cos^2x+cosx=cos2x Solve algebraically for exact solutions in the interval [0,2π). Use your grapher only to support your algebraic work. 24. -
## Math - Trig - Double Angles

Prove: sin2x / 1 - cos2x = cotx My Attempt: LS: = 2sinxcosx / - 1 - (1 - 2sin^2x) = 2sinxcosx / - 1 + 2sin^2x = cosx / sinx - 1 = cosx / sinx - 1/1 = cosx / sinx - sinx / sinx -- Prove: 2sin(x+y)sin(x-y) = cos2y - cos2x My Attempt: RS: = 1 - 2sin^2y - 1 - -
## Trigonometry

solve Sinx=Cos2x-1 for 0¡Üx<2¦Ð -
## math

solve for sinx=cos2x (between 0 and 2pi) -
## Trig Help

Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1-cos^2x)]/[sinx+1] =??? This is where I'm stuck. Can someone help me. Please check what I got is -
## trig

how do you solve: find cos2x if sinx is equal to 1/5 -
## Trig

Solve Sinx=Cos2x-1 for all values between 0 and 2pi -
## maths

solve for x for angles between o and 360 degrees (i)cos2x+sinx=0 -
## Calculus grade 12

solve for (<_ = less than or equal to / pie = pie sign / -pie = negative pie) 3 sin²x = cos²x ; 0 <_ x < 2pie cos²x - sin²x = sinx ; -pie < x <_ pie -
## Trig

If sinx = 1/4 Find the exact value of cos2x I used the formula for double angle identities : cos2x = 1-2sin^2x and got the answer 1/2 The book says the answer is 7/8 How would they get to that conclusion? -
## Math please help quick

Which of the following are identities? Check all that apply. (Points : 2) sin2x = 1 - cos2x sin2x - cos2x = 1 tan2x = 1 + sec2x cot2x = csc2x - 1 Question 4. 4. Which of the following equations are identities? Check all that apply. (Points : 2) Question 5. -
## Advanced Math

Could someone help me with those two questions. I need to know how to do them without calculator solve for the interval [0,2pie] 1) cos^2*x-cos2x=0 2) tan(x/2)-sinx=0 -
## trig

sinx = 4/5 and x terminates in Quadrant II Find sin2x and cos2x How to get the answers, which are sin2x = -24/25, cos2x = -7/25? -
## Calculus

Solve identity, (1-sin2x)/cos2x = cos2x/(1+sin2x) I tried starting from the right side, RS: =(cos²x-sin²x)/(1+2sinxcosx) =(cos²x-(1-cos²x))/(1+2sinxcosx) and the right side just goes in circle. May I get a hint to start off? -
## Trig.

Can someone please help me with this? Find all solutions to the equation in the interval [0,2pi) cos2x=sinx I know I have to use some sort of identity, but I have no idea how to go about to solve this. -
## Math (Trig) a.s.a.p, thanks!

verify the identities: 1.) 2tanx-(1+tanx)^2 = -secx 2.) cos(x-330degrees)=1/2(ã3cosx-sinx) *the ãis only over the 3* 3.) 1+cos2x/sinx=cotx any help is greatly appreciated!! -
## Trig

Verify the identity: tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(2cos^2 x -1) =sinx(2cos^2 x - 1)/cosx Right Side = 2sinx cosx - sinx/cosx =(2sinxcos^2 x - sinx)/cosx =sinx(2cos^2 x -1)/cosx = L.S. Q.E.D. -
## math

solve each equation for 0=/<x=/<2pi sin^2x + 5sinx + 6 = 0? how do i factor this and solve? 2sin^2 + sinx = 0 (2sinx - 3)(sinx +2) sinx = 3/2, -2 how do I solve? These are not part of the special triangles. 2cos^2 - 7cosx + 3 = 0 (2cosx - 1)(cosx - -
## maths please answer NOW!!!!!

solve algebreically the equation cos2x=1-sinx also cosx=cosx-1 and please also sin2x=3sinx -
## calc

Where do I start to prove this identity: sinx/cosx= 1-cos2x/sin2x please help!! Hint: Fractions are evil. Get rid of them. Well, cos2x = cos2x - sin2x, so 1-coscx = 1 - cos2x - sin2x = 1 - cos2x + sin2x You should be able to simplify this to 2*something -
## Math

prove identity Sin2x - sinx/cosx + cos2x= sinx/cosx + 1 -
## trig

how does this work? (Sin2x+cos2x+2sin²x)/(sinx+cosx) = (sinx+cosx -
## Trig

Prove the following functions: (sinx+sin2x)/(1+cosx+cos2x)=tan x (cos3x/sinx)+(sin3x/cosx)=2cot2x tan2x=(2/cotx-tanx) theses are due in the am. Fastness would be great? -
## Math

Solve the equation 2cos2x = √3 for 0°≤x≤360° I did this: cos2x = √3 /2 2x=30 x=15 x=15, 165, 195, 345 Is this correct? Solve the equation √3 sin2x + cos2x = 0 for -π≤x≤π No idea how to approach this one Thanks a bunch for the help! -
## Trig........

I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top and bottom by 1+sinx, -
## Math/Calculus

Please take a look at my work below and provide a good critique: Solve the differential equation using the method of undetermined coefficients or variation of parameters. y'' - 3y' + 2y = sin(x) yc(x)= c1*e^2x+c2*e^x y"-3y'+2y=sin(x) r^2-3r+2=0 -
## Trigonometry

Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do. Simplify #3: -
## Math 12

Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do. Simplify #3: -
## trig

Find the general solution for x if cos2x + sin3x = sinx -
## Calculus

In the interval (0 is less than or equal to x which is less than or equal to 5), the graphs of y=cos(2x) and y=sin(3x) intersect four times. Let A, B, C, and D be the x-coordinates of these points so that 0<A<B<C<D<5. Which of the definite -
## Math

Use a double-angle formula to find the exact value of cos2x when sinx = 3/5, where pi/2 < x < pi. -(7/25) or 7/25? -
## Math

Find the exact solution of cos2x+sinx=0 in the interval [0,2pi) -
## Identities

Verify the following identity cos2x/sin2x+sinx/cosx=csc2x -
## Trig

help please!: (sinx + sin2x +sin3x) / (cosx + cos2x + cos3x) = tan2x -
## trig

Find the solutions to the equation cos2x=sinx and 2cos^2x-3cosx-3=0 -
## Pre Calculus

Establish the following identity sin 2x/sinx - cos2x/cosx = secx -
## Trigonometry

find sin2x, cos2x, and tan2x if sinx= -2/sqrt 5 and x terminates in quadrant III -
## Trig

solve to four decimals places 5.0118 sinx - 3.1105=0 :for all real x values I will be happy to critique your work. Solve for sinx, then arcsinx -
## calculus

Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the derivative of sin[sinx]. And -
## Trig

solve to four decimal places 1.3224sinx + 0.4732=0 :for all real x values solve for sinx, then arcsinx. solve for sinx, then arcsinx. so its sinx= -.4732/1.3224 and i get -.36594 but how do you get the other values for a real x Ok, use your calculator. -
## trig

I keep trying to find the power reducing formula for sin^4(x), but I can't seem to get all the fractional parts correct. The answer I should be getting is: sin^4(x)=(1/8)cos4x-(1/2)cos2x+(3/8) I can only get this far knowing feeling confident. When I go -
## Trigonometry

Simplify the expression using trig identities: 1. (sin4x - cos4x)/(sin2x -cos2x) 2. (sinx(cotx)+cosx)/(2cotx) -
## Trig.......

I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr Reiny I'll tell my teacher -
## Math

Solve the integral from [0,pi] of x*sinx*cosx*dx I don't understand how to solve this problem because if I let u =x and du =1 and dv=sinx*cosx, I can't find v. How do you solve this? -
## Math (Integrals) (Using given method)

I would like to solve the ∫sin^2(pix) dx Using the given substitution identity: sin^2(x) = (1/2)(1-cos2x) This is what I did so far: ∫sin^2(pix) let u = pix du = pi dx (1/pi)∫sin^2(u)du Applying the identity is where I'm lost on how to continue -
## Pre-Calculus

Find all solutions to the equation in the interval [0, 2pi) cos4x-cos2x=0 So,this is what i've done so far: cos4x-cos2x=0 cos2(2x)-cos2x (2cos^2(2x)-1)-(2cos^2(x)-1) No idea what to do next. -
## Trig Help

Given that cos2x=7/12 and "270 equal or < 2x equal or < 360", find sinx. Please help and Thank you -
## Pre calc

Please help me proof the identity. This isn't even math to me, I can not understand it so please show all work. God bless A)1-cos2x/sin2x=tanx B) sin4x/sinx=4cosxcos2x -
## Trigonometry

I can't remember what formula to use for these. Note- the X's may be replaced by a theta symbol. I just didn't have one on my keyboard. Use the given information to find sin2X, cos2X and tan2X if 0< X <pi/2 1. sinX=12/13 2. cosX=3/5 3. tanX=2/3 -
## Trigonometry.

( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated side and try to change it to -
## math

sin*sin=1.732*sinx*cos2x -
## Precal

show that: (1-tanx)/(1+tanx)= (1-sinx)/(cos2x) -
## trigonometry

can i use factoring to simplify this trig identity? the problem is sinx + cotx * cosx i know the answer is cscx and i know how to get it but i want to know if i can do factoring to get it bc i tried to but it wont give me the answer . this is the step i -
## Pre-cal

1) Solve the equation sinx(sinx+1) = 0 I am having trouble with the first step. Could I do sin x^2 + 1 =0? -
## math

solve sinx/2,cosx/2 & tanx/2 of the following:- (a)sinx=1/4 & x lies in second quadrant. -
## Math C30

Equations containing circular functions and unit circle... solve 2cosx sinx + sinx = 0 -
## Mathj C30

Solving Equations Containing Circular Functions : Solve 2cosx sinx + sinx = 0 when x is between 0 and 2pi -
## trig

solve equation for exact solution if possible leave answer in degree sinx/2=square root of 2- sinx/2 -
## maths - trigonometry

I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the second. [sinx + tanx]/[cosx + 1] -
## Maths

If y=Sinx^Sinx^Sinx.......infinity then prove that dy/dx=2ycotx/1-ylog(sinx).? -
## Simplifying with Trigonometry Identities

Hi, I am a senior in High School having a really difficult time with two problems. I have to prove using the trigonometric identities that they equal each other but I am having a really hard time trying to get them to equal each other. I've spent over 2 -
## Trigonometry Check

Simplify #3: [cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] = [cosx-((1)cosx-(0)sinx)sinx]/[cosx-((-1)cosx+(0)sinx)tanx] = [cosx-cosxsinx]/[cosx+cosxtanx] = [cosx(1-sinx]/[cosx(1+tanx] = -
## Mathematics - Trigonometric Identities

Prove: (tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx) What I have so far: L.S. = (sinx / cosx) sinx / (sinx / cosx) + sinx = (sin^2x / cosx) / (sinx + (sinx) (cosx) / cosx) = (sin^2x / cosx) / (cosx / sinx + sinxcosx) -
## Math

1) evaluate without a calculator: a)sin(3.14/4) b) cos(-3(3.14)/4) c) tan(4(3.14)/3) d) arccos(- square root of three/2) e) csctheata=2 2) verify the following identities: a) cotxcosx+sinx=cscx b)[(1+sinx)/ cosx] + [cosx/ (1+sinx)]= 2secx c) (sin^2x + -
## Pre-calc

prove the identity: (cosx)(tanx + sinx cotx)=sinx+cos(squared)x i need steps to show how i got the answer generally, it is a good idea to change all trig ratios to sines and cosines, and start with the more complicated-looking side. so.... LS = -
## Math - Pre- Clac

Prove that each of these equations is an identity. A) (1 + sinx + cos x)/(1 + sinx + cosx)=(1 + cosx)/sinx B) (1 + sinx + cosx)/(1 - sinx + cosx)= (1 + sin x)/cosx Please and thankyou! -
## drwls

My previous question: Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity. (secx/sinx)*(cotx/cscx) = (secx/cscx)(cotx/sinx) = (sinx/cosx)*cotx*(1/sinx) "The last steps should be obvious" Not to me. I can convert (sinx/cosx) to tanx if that's even what -
## Math (trigonometric identities)

I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx - sinx - cos^3x/Cosx = -
## Math trignometry

how can you condense these cos^3x* cos2x^2 cos^3x+ cos2x^2 cos^3x- cos2x^2 Thanks a lot :) -
## your preciouss helpp please

how can you condense these cos^3x* cos2x^2 cos^3x+ cos2x^2 cos^3x- cos2x^2 Thank you soo much -
## Math

Im really struggling with these proving identities problems can somebody please show me how to do these? I'm only aloud to manipulate one side of the equation and it has to equal the other side of the equation at the end Problem 1. Sinx/(cotx+1) + -
## calc

what is the derivative of sinx- cosy = 0 in its simplest form? I got to the point y'=-consx/siny, but I wasn't too sure. This could probably simplified further. help! You're supposed to treat y as a function of x and differentiate it implicitly. We have -
## Trigonometry

4. Find the exact value for sin(x+y) if sinx=-4/5 and cos y = 15/17. Angles x and y are in the fourth quadrant. 5. Find the exact value for cos 165degrees using the half-angle identity. 1. Solve: 2 cos^2x - 3 cosx + 1 = 0 for 0 less than or equal to x -
## Maths

Prove that cos3x/cosx-cos6x/cos2x=2(cos2x-cos4x) -
## Trig identies, Calculus

Use the identities cos^2 x + sin^2 x =1 and cos2x=cos^2 x -sin^2 x to show that cos^4 x -sin^4 x = cos2x Im not sure how, I can solve my problem with half angle identities but im not sure where to start with this. -
## trig

tanx = 5/12 and sinx<0 Find sin2x and cos2x How to get the answers of sin2x = 120/169 and cos 2x = 119/169? -
## Trigonometic

Solving Trigonometic Equations solve for x and give the answers as a equations : ( by radian) 1)cos(sinx)=1 <<<and thanks >>> We know sin 2x = 2(sinx)(cosx) so (sinx)(cos)=1/2(sin 2x) So we can change your equation from (sinx)(cosx)=1 to -
## Mathematics - Trigonometric Identities

Prove: sinx + tanx = tanx (1 + cosx) What I have so far: LS: = sinx + tanx = sinx + (sinx / cosx) = (sinx) (cosx) + sinx / cos = tanx (cosx + sinx) I don't know what to do now -
## math

tanx+secx=2cosx (sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 = 2(1-sin^2x) 2sin^2x + sinx-1=0 (2sinx+1)(sinx-1)=0 x=30 x=270 but if i plug 270 back into the original equation i get undefined -
## Calculus

Q: If y=sinx/(1+tanx), find value of x not greater than pi, corresponding to maxima or minima value of y. I have proceeded thus- Equating dy/dx=0 we get{ (1+tanx)cosx-sinx.sec^2 x}/(1+tanx)^2=0……..(A) Or cosx+sinx=sinx.sec^2 x or cosx= sinx.sec^2-sinx -
## Math

I'm only aloud to manipulate one side of the problem and the end result has to match the other side of the equation Problem 1. sinx + cosx + sinx + tanx + cosxcotx = secx + cscx Problem 2. ((sinx + cosx)/(1 + tanx))^2 + ((sinx - cos^2x)/(1 - cotx))^2 = 1 -
## trigonometry

how do i simplify (secx - cosx) / sinx? i tried splitting the numerator up so that i had (secx / sinx) - (cosx / sinx) and then i changed sec x to 1/ cosx so that i had ((1/cosx)/ sinx) - (cos x / sinx) after that i get stuck -
## advanced functions

Use the Pythagorean identity to show that the double angle formula for cosine can be written as a) cos2x = 1 - 2sin^2x b) cos2x = 2cos^2x - 1 -
## Math

How do I solve this? tan^2x= 2tanxsinx My work so far: tan^2x - 2tanxsinx=0 tanx(tanx - 2sinx)=0 Then the solutions are: TanX=0 and sinX/cosX = 2 sin X Divide through by sinX: we have to check this later to see if allowed (ie sinX is not zero) cosX=1/2 -
## Calculus

Find the derivative of y with respect to x: y=(1+cos²x)^6 y'=6(1+cos²x)^5 How do you derive inside the brackets? The answer says -sin2x, but wouldn't it be -2sinx, using the chain rule. -
## Pre-Calc

Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx + cosx/cosx)/(sinx/cos x - 1/cosx + cosx/cosx)= tanx + cosx (just working on the left side) ((sinx + 1 - cosx)/cosx)/((sinx - 1 + -
## Trig

#1 (1/sinx)(sin^2x + cos^2x(sinx/cosx) )/(sinx + cosx) = (sinx + cosx)/(sinx+cosx) = 1 Reiny helped with this question but I don't how Reiny did it. Please explain -
## Calculus

If y=3/(sinx+cosx) , find dy/dx A. 3sinx-3cosx B. 3/(sinx+cosx)^2 C. -3/(sinx+cosx)^2 D. 3(cosx-sinx)/(sinx+cosx)^2 E. 3(sinx-cosx)/(1+2sinxcosx) -
## Trig

Simplify: (cosx + cos2x + cos3x + cos4x + cos5x + cos6x) / (sinx + sin2x + sin3x + sin4x + sin5x + sin6x) into a single Cotangent function. Using the sum-to-products, I was able to get remove some of the addition in attempts to get full multiplication, but -
## CALCULUS!!!

case 1. Find the absolute maxima and minina values of f(x)=sin2x+cos2x on [0, pie] Specify both the x and y coordinateof the absolute maximum and absolute minimum. Gives answers in exact form. case 2. Use the second derivative test for relative maxima nd -
## Math-gr12

How would I solve this equation for x in the interval 0<x<2pi tan^2x sinx - sinx/3 = 0 Please explain it step by step..ive been stuck on this question for far too long..sigh -
## Trig

Simplify sin x cos^2x-sinx Here's my book's explanation which I don't totally follow sin x cos^2x-sinx=sinx(cos^2x-1) =-sinx(1-cos^2x) =-sinx(sin^2x) (Where does sine come from and what happend to cosine?) =-sin^3x -
## math

Solve for: 1+cosx/sinx + sinx/1+cosx = 1 My teacher said no solution, but I don't know how he got that. -
## Pre-Calc

Radical ((1-sinx)/(1+sinx))= (1-sinx)/cosx (absolute) Simplify -
## Math

Differentiate (1- sinx) / (1 +sinx) How would I do this? The answer is (-2cosx)/ (1 + sinx)^2 Thanks in advance :)!! -
## math

y= 2sin^2 x y=1- sinx find values of x inthe interval 0<x<360 if 2sin^x = 1-sinx this can be arranged into the quadratic. 2sin^2 x + sin x -1=0 (2sinx -1)(sinx + 1)=0 or sinx = 1/2 and sin x=-1