solving the equations?

Question

solve the equation on the intervel of 0 < equal to theta < 2 pi

a) cos^2 theta - sin ^2 theta + sin theta = 0

b) cos (2 theta) + 4 cos theta +3 = 0

please help me and explain this step by step, thanks

Answer 1

cos^2 + sin^2 = 1
cos^2 = 1 - sin^2

1 - sin^2 - sin^2 +sin = 0

Now replace sin(theta) with x:

-(2x^2 -x -1) = 0
Look familiar??

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

So x = 1, -1/2

And since x = sin(theta), replace it again...

sin(theta) = -1/2
theta = 11 pi/6

sin(theta) = 1
theta = pi/2

a) So theta = pi/2 or 11 pi/6


cos (2theta) = 2 cos^2(theta) -1

2cos^2 + 4 cos -1 +3= 0

Replace cos with x:
2x^2 +4x +2 = 0

x^2 +2x + 1 = 0
(x+1)^2 = 0

x = -1

replace x with cos:
cos (theta) = -1
theta = pi

b) So theta = pi

Answer 2

a)
cos² x - sin² x + sin x = 0
(1 - sin² x) - sin² x + sin x = 0
1 - 2sin² x + sin x = 0
2sin² x - sin x - 1 = 0
(2sin x + 1)(sin x - 1) = 0
2 sin x + 1 = 0
sin x = -1/2
x = 7π/6 or x = 11π/6
sin x - 1 = 0
sin x = 1
x = π/2

b)
cos 2x + 4cos x + 3 = 0
cos² x - sin² x + 4cos x + 3 = 0
cos² x - (1 - cos² x) + 4cos x + 3 = 0
2cos² x + 4cos x + 2 = 0
cos² x + 2cos x + 1 = 0
(cos x + 1)² = 0
cos x + 1 = 0
cos x = -1
x = -3π/2