Some trigonometry equations

a) 4sin x  =  cosx        square both sides

16sin^2x  =  cos^2x

16sin^2x  = 1 -  sin^2x

17sin^2x  = 1

sin^2x   = 1/17               take both roots

sinx  = ±√ (1/17)

To find both values in degrees...take the sine inverse

arcsin (√[1/17])  ≈  14.04°

arcsin (-√ [ 1/17] )  ≈ -14.04°.......however.....we must   subtract this from -180° to get the correct answer

-180 - - 14.04  ≈  -165.96°

Here's a graph that shows the intersection of these curves : https://www.desmos.com/calculator/c0dyooszqg

b)  sin2x  = cos (pi/2  - x)     

Note   cos(pi/2 - x)   =  sin x

So we have

sin2x  = sinx

2sinxcosx  =  sinx

2sinx cosx  - sinx  = 0

sinx ( 2cosx - 1)  = 0

So...either   sin x  = 0    and this happens  at    0,  pi  and 2pi

Or

2cosx - 1  = 0

cosx  =  1/2   and this happens at     pi/3   and 5pi/3

Here's a graph :  https://www.desmos.com/calculator/y1o4mppidr

Thank you so much! This really helped me! :D

In a) I didn't think of squaring both sides to get cos^2x = 1 -  sin^2x, so thank you!

In b) I didn't think of that cos(pi/2 - x)   =  sin x !!! So thank you, this is what helped me the most.

Thanks!!! Much love from Sweden.

Tack  !!!