Solving Trig Equations?

Question

0 <= theta < 2pi

Could someone please explain how to solve these problems? Thanks.

1. 4 cos^2 theta - 3 = 0

2. cos^2 theta - sin^2 theta + sin theta = 0

Answer 1

1.)
4cos(A)^2 - 3 = 0
4cos(A)^2 = 3
cos(A)^2 = (3/4)
cos(A) = sqrt(3)/2
A = pi/6, (5pi/6), (11pi/6), or (7pi/6)

2.)
cos(A)^2 - sin(A)^2 + sin(A) = 0
(1 - sin(A)^2) - sin(A)^2 + sin(A) = 0
1 - sin(A)^2 - sin(A)^2 + sin(A) = 0
-2sin(A)^2 + sin(A) + 1 = 0
2sin(A)^2 - sin(A) - 1 = 0
(2sin(A) + 1)(sin(A) - 1) = 0

2sin(A) + 1 = 0
sin(A) = (-1/2)
A = (11pi/6) or (7pi/6)

sin(A) - 1 = 0
sin(A) = 1
A = pi/2

ANS : (pi/2), (11pi/6), or (7pi/6)

Answer 2

1. 4 cos^2 theta - 3 = 0
cos^2 theta =3/4
cos theta = √3/2
theta = 30°

Answer 3

2. 1 - sin^2 a - sin^2 a + sin a = 0
2sin^2 a - sin a -1 = 0
(2sin a +1)(sin a -1) = 0
sin a = -1/2 or 1

a = 210 or 330 or a = 90 deg

Answer 4

4cos²θ - 3 = 0
4cos²θ = 3
cos²θ = 3/4
cosθ = ±sqrt(3)/2
θ = {π/6, 5π/6, 7π/6, 11π/6}

cos²θ - sin²θ + sinθ = 0
1 - 2sin²θ + sinθ = 0
2sin²θ - sinθ - 1 = 0
2sin²θ - 2sinθ + sinθ - 1 = 0
2sinθ(sinθ-1)+1(sinθ-1) = 0
(2sinθ+1)(sinθ-1) = 0
sinθ = {-1/2, 1}
θ = {π/2, 7π/6, 11π/6}