# equations trig

1) ctgx+cosx=1+ctgx*cosx

2) sin^4x+cos^4x=1/2

3) (sinx-cosx)^2=sin2x

4) sin^4x-cos^4x=7/2(sinxcosx)

5) sin^2x-sinxcosx+cosx-sinx=0

1. 👍
2. 👎
3. 👁
1. How about you try first and post your results.

1. 👍
2. 👎
2. 2) sin^4x+cos^4x=1/2
sin^2x*sin^2x+cos^4x=1/2
(1-cos^2x)(1-cos^2x)+cos^4x=1/2
1-cos^2x-cos^2x+2cos^4x=1/2
2cos^4x-2cos^2x+1=1/2
2cos^4x-2cos^2x+1/2=0
x=pi/4+2kpi x2=7pi/4+2kpi

1. 👍
2. 👎

## Similar Questions

1. ### math;)

The equation 2sinx+sqrt(3)cotx=sinx is partially solved below. 2sinx+sqrt(3)cotx=sinx sinx(2sinx+sqrt(3)cotx)=sinx(sinx) 2sin^2x+sqrt(3)cosx=sin^2x sin^2x+sqrt(3)cosx=0 Which of the following steps could be included in the

2. ### Math

Suppose f(x) = sin(pi*cosx) On any interval where the inverse function y = f –1(x) exists, the derivative of f –1(x) with respect to x is: a)-1/(cos(pi*cosx)), where x and y are related by the equation (satisfy the equation)

3. ### 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

4. ### Math

verify: sin^3 x - cos^3 x = (sinx+cosx)(1-sinxcosx)

1. ### 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

2. ### 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

Which of the following are trigonometric identities? (Can be more then one answer) tanx cosx cscx = 1 secx-cosx/secs=sin^2x 1-tanxtany=cos(x+y)/cosxcosy 4cosx sinx = 2cosx + 1 - 2sinx Find all solutions to the equation cosx

4. ### Trig Identities

Prove the following identities: 13. tan(x) + sec(x) = (cos(x)) / (1-sin(x)) *Sorry for any confusing parenthesis.* My work: I simplified the left side to a. ((sinx) / (cosx)) + (1 / cosx) , then b. (sinx + 1) / cosx = (cos(x)) /

1. ### maths

(Sin^3x-cos^3x)/(sinx-cosx) – cosx/sqrt(1+cot^2x)-2tanxcotx=-1 where x∈(0,2pi) general value of x.

2. ### Math help again

cos(3π/4+x) + sin (3π/4 -x) = 0 = cos(3π/4)cosx + sin(3π/4)sinx + sin(3π/4)cosx - cos(3π/4)sinx = -1/sqrt2cosx + 1/sqrt2sinx + 1/sqrt2cosx - (-1/sqrt2sinx) I canceled out -1/sqrt2cosx and 1/sqrt2cosx Now I have 1/sqrt sinx +

3. ### Trigonometry

verify the identity: (sin^3x-cos^3x)/(sinx-cosx)=1+sinxcosx

4. ### Trig

If secx = 8 and -pi/2 < x < 0, find the exact value of sin2x Use the identity sin 2x = 2(sinx)(cosx) if secx = 8, then cosx = 1/8 where x is in the fourth quadrant. consider a right angled triangle with x=1, r=8, then y=?? by