Trig Equations?

Question

0 <= theta < 2pi

Could someone please explain how to solve these? Thanks.

1. sin 2theta sin theta = cos theta

2. cos 2theta + 5 cos theta + 3 = 0

Answer 1

if by these you mean

1.)
sin(2A)sin(A) = cos(A)
2cos(A)sin(A)^2 = cos(A)
2sin(A)^2 = 1
sin(A)^2 = 1/2
sin(A) = sqrt(2)/2
A = pi/4 or (5pi/4)

2.)
cos(2A) + 5cos(A) + 3 = 0
2cos(A)^2 - 1 + 5cos(A) + 3 = 0
2cos(A)^2 + 5cos(A) + 2 = 0
(2cos(A) + 1)(cos(A) + 2) = 0

2cos(A) + 1 = 0
2cos(A) = -1
cos(A) = (-1/2)
A = (2pi/3) or (4pi/3)

cos(A) + 2 = 0
cos(A) = -2
this one isn't possible

Answer 2

Usually, trial and error is as good a way as any.
For example, in #2, you can treat "cos theta" as the variable and solve a quadratic for two roots of cos theta, but to get a numerical answer, would require you to track down the result either on a calculator or a trig table.

Answer 3

0 ≤ Θ < 2π

1. sin 2Θ sin Θ = cos Θ

(2sinΘcosΘ)sin Θ = cos Θ

(2sinΘcosΘ)sin Θ - cos Θ = 0

cosΘ(2sin²Θ - 1) = 0

cosΘ = 0 or 2sin²Θ - 1 = 0

cosΘ = 0 or sinΘ = ± (√2)/2

Θ = π/2, 3π/2 or π/4, 3π/4, 5π/4, 7π/4

Θ = π/4, π/2, 3π/4, 5π/4, 3π/2, 7π/4

2. cos 2Θ+ 5 cos Θ + 3 = 0

2cos²Θ - 1 + 5 cos Θ + 3 = 0

2cos²Θ + 5 cos Θ + 2 = 0

(2cosΘ + 1)(cosΘ + 2) = 0

2cosΘ + 1 = 0 or cosΘ + 2 = 0

cosΘ = -1/2 or cosΘ = -2

Θ = 2π/3, 4π/3 or no solution

Θ = 2π/3, 4π/3