2sin(theta) + cos(theta) -1 = 0?

Question

solve between 0 and 360 degrees.

Answer 1

2sinθ = 1 - cosθ

(2sinθ)² = (1 - cosθ)²

4sin²θ = 1 - 2cosθ + cos²θ
4(1-cos²θ) = 1 - 2cosθ + cos²θ
4 - 4cos²θ = 1 - 2cosθ + cos²θ
5cos²θ - 2cosθ - 3 = 0

(5cosθ+3)(cosθ -1) = 0

cosθ = -3/5 or cosθ = 1
θ = arccos(-3/5), 360°-arccos(-3/5), 0°
θ = 126.9°, 233.1°, 0°
Now, because we used the technique of squaring both sides of an equation, we have to check our answers in the original equation. Most likely something won't check:

126.9° checks
233.1° does not check
0° checks

Answer 2

first let´s call theta =t and try to put
2sin r+cos t = Rcos(t+a) =R (cost*cos a-sint*sin a)
so
Rcos a=1
R sin a=-2
square and sum R^2= 5 and R = sqrt5
tan a= -2 so a=-1.11rad so the equation becomes
sqrt5 *cos(t-1.11)=1
cos(t-1.11)=sqrt5/5 =0.45
t-1,11=1.11 and t-1.11 = 5.18
t= 2.22 rad and 6.29 rad(apprx).If you inspect the equation you see that t= 0 =2pi is solution
so the solutions are t=0 or 2pi and t=2.22 radians

Answer 3

2sin(θ) + cos(θ) - 1 = 0
2sin(θ) - 1 = - cos(θ)
2sin(θ) - 1 = -√(1 - sin²(θ))
(2sin(θ) - 1)² = 1 - sin²(θ)
4sin²(θ) - 4sin(θ) + 1 = 1 - sin²(θ)
5sin²(θ) - 4sin(θ) = 0
sin(θ) (5sin(θ) - 4) = 0
So sin(θ) = 0 for θ=0 and θ=180, and
(5sin(θ) - 4) = 0 for θ = sin^-1 (4/5)
But θ=180 doesn't work in the original equation, so the answers are
θ=0, sin^-1 (4/5)
evaluate this second term for the first and second quadrants and only keep the value(s) that work.