Question

Question #bb33b

Answer 1

(4sinx)^2 = (cosx)^2#

`16sin^2x = cos^2x`

`16sin^2x = 1- sin^2x`

`17sin^2x = 1`

`sin^2x = 1/17`

`sinx= +-sqrt(1/17)`

`x = 14˚`

Clearly, this is the only solution that will work, since we need a solution in the first quadrant, where sine and cosine are positive.

Hopefully this helps!