solve the equation

Question

1. cos(θ) − sin(θ) = 1

2.2 cos(θ) tan(θ) + tan(θ) = 1 + 2 cos(θ)

3. sin(θ) cos(3θ) + cos(θ) sin(3θ) = 0

4. sin(2θ) cos(θ) − cos(2θ) sin(θ) = 1/2

5. cos(2θ) + cos(θ) = 2

6. cos(2θ) + sin2(θ) = 0

Answer 1

#1, look how I did this similar problem, and follow the same steps

http://www.jiskha.com/display.cgi?id=1461697741

#2
2 cos(θ) tan(θ) + tan(θ) = 1 + 2 cos(θ)
2cosθtanθ - 2cos + tanθ - 1 = 0
2cosθ(tanθ - 1) + (tanθ -1) = 0
by grouping:
(tan - 1)(2cosθ + 1) = 0
tanθ = 1 or cosθ = -1/2
θ = 45 or θ = 225 or θ = 240 or θ =300 degrees
in radians: θ = π/4, 5π/4, 4π/3, 5π/3

#3 and #4 are based on the formula
sin(A ± B) = sinAcosB ± cosAsinB

I will do #4, you do #3
sin(2θ) cos(θ) − cos(2θ) sin(θ) = 1/2
sin(2θ - θ) = 1/2
sin θ = 1/2
θ = 30 degrees or θ = 150 degrees
or θ = pi/6 or 5pi/6

#5
I will assume you mean:
cos(2θ) + sin^2 (θ) = 0
cos^2 θ - sin^2 θ + sin^2 θ = 0
cos^2 θ = 0
cosθ = 0
θ = 90 or θ = 270 degrees
θ = pi/2 or 3pi/2