Question

how do you solve Sin2xSinx -cos x=0?

Answer 1

Since sin(2x) = 2 sinx cosx, we have

 2 sinx cosx sinx - cosx = 0

 cosx (2 sin²x - 1) = 0

Therefore

 cosx = 0 or 2 sin²x - 1 = 0

The solutions of cosx = 0 are x = any whole number of rightangles.

For 2 sin²x - 1, we get

 sinx = 1/√2 or -1/√2

so x = any odd multiple of a half-rightangle, i.e. 45°, 135°, 225° etc, or, in radians,

 x = π/4, 3π/4, 5π/4 etc

Answer 2

cos x=0

x=pi/2, 3pi/2

sin x =+/- 1/sqrt2

x=pi/4, 3pi/4, 5pi/4, 7pi/4