Solve over the interval [ 0 degrees , 360 degrees ). sin^2x * cosx = cosx

Question

what I did for this question is move cosx to the other side of the equal sign to get sin^2x * cosx - cosx = 0 ... then cosx(sin^2x-1)=0 ... Cosx= 0 & sin^2x=1 , im unsure of what to do from here?

Answer 1

sin^2(x) = 1

sinx =+ or - sqr1 = + or -1

x = 90 or 270 degrees

cosx = 0

x= 90 or 270 degrees

Answer: x = 90 or 270 degrees or pi/2, or 3pi/2 radians

If you don't recognize how to get from cosx=0 to x=90, and 270, draw a unit circle, the more vertical the angle, the closer to 90 or 270, the smaller cosx is, until it reaches zero. or punch 0 into a calculator and then take the inverse cosine. Same with sinx=1. Use a calculator that has inverse trig functions., inverse sine and inverse cosine.

IF by chance, the problem really read sin2x and not sin^2x, the answers include more angles.

Answer 2

cosx(sin²x -1) = 0; - cos³x = 0, cosx = 0; x = 90° or x = 270°.

Answer: x = 90°, x = 270°