1. Given, tan2x + tan x = 0 Let tan x be a Thus, a^2 + a = 0  a = 0, -1 Thus, tan…

Let tan x be a

Thus, a^2 + a = 0

 a = 0, -1

Thus, tan x = 0 => x = 0
And tan x = -1 => x = 135o = 3π/4

Thus the solutions are: {0, 3π/4}

2.
Given,
2cos2 (2θ) = 1- cos 2θ

Let cos 2θ = a. We can write this expression as,

2a2 + a – 1 = 0
 (2a – 1)(a + 1) = 0
 a = ½ , -1

Thus, cos 2θ = ½ , -1

When cos 2θ = ½
 2θ = 60, 300
 θ = 30, 150 = π/6, 5π/6

Also, when cos 2θ = -1
 2θ = 180
 θ = 90 = π/2