Trig (Last URGENT)
posted by Anonymous .
sin2x+cosx=0 , [180,180)
= 2sinxcosx+cosx=0
= cosx(2sinx+1)=0
cosx=0
x1=cos^1(0)
x1=90
x2=36090
x2=270
270 doesn't fit in [180,180) what do I do? Or maybe I did something wrong.
sinx=1/2
x1=sin^1(1/2)
x1=30
x2=18030
x2=150
is this correct?
Please and Thank you

Trig (Last URGENT) 
Reiny
but your domain is from 180 to 180 , so even though 270 will work in the equation , it is beyond your domain.
But, by looking at the cosine graph , you will see that it also has an xintercept at x = 90°
so for your first one,
x = 90, 0, 90
the second part for sinx = 1/2
your answers of 30° or 150° are correct for the given domain.
Respond to this Question
Similar Questions

PreCalc
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 … 
precalc
Solve: cos(2x180)  sin(x90)=0 my work: cos2xcos180 + sin2xsin180= sinxcos90  sin90cosx cos2x  sin2x= cosx cos^2x + sin^2x 2sinxcosx=cosx I'm stuck here. I tried subtracting cosx from both sides and making sin^2x into 1 cos^2x, … 
Trig.......
I need to prove that the following is true. Thanks (2tanx /1tan^x)+(1/2cos^2x1)= (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 … 
Trig........
I need to prove that the following is true. Thanks (cosx / 1sinx ) = ( 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/(1sinx) multiply top and … 
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(1sin^2x) 2sin^2x + sinx1=0 (2sinx+1)(sinx1)=0 x=30 x=270 but if i plug 270 back into the original equation … 
trig
Solve for [0, 360) 2sinxcosx + cosx =0 2sinxcosx = cosx 2sinx = cosx/cosx sinx = 1/2 {210, 330) Is this correct? 
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? 
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? 
Trigonometry Check
Simplify #3: [cosxsin(90x)sinx]/[cosxcos(180x)tanx] = [cosx(sin90cosxcos90sinx)sinx]/[cosx(cos180cosx+sinx180sinx)tanx] = [cosx((1)cosx(0)sinx)sinx]/[cosx((1)cosx+(0)sinx)tanx] = [cosxcosxsinx]/[cosx+cosxtanx] = [cosx(1sinx]/[cosx(1+tanx] … 
Precalculus/Trig
I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1  cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1cosx Simplified: cosx + sin^3x/sin^3x = cscx/1cosx I don't know …