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.

24,18t

step-by-step explanation: