Solve the following trigonometric equation for 0to2pi

(a) cos (4x) - 9 sin (2x) + 4 = 0

x = pi/12, 5pi/12, 13pi/12, 17pi/12.

Step-by-step explanation:

Note that cos (4x) = 1 - 2sin^2 (2x)

Substituting we have:

1 - 2sin^2 (2x) - 9 sin (2x) + 4 = 0

2 sin^2 (2x) + 9 sin (2x) - 5 = 0

(2 sin 2x - 1) (sin 2x + 5) = 0

sin 2x = 1/2 and sin 2x = - 5.

sin 2x = 1/2, gives 2x = pi/6, 5pi/6, 13pi/6, 17pi/6.

There are no solutions to sin 2x = - 5 because the range of sin x is - 1 to + 1.

So x = pi/12, 5pi/12, 13pi/12, 17pi/12.