What is the general solution for cos(2θ)+sin(θ)=1 ?

The general solution for cos(2θ)+sin(θ)=1 is θ=2pin,θ=pi+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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cos(2θ)+sin(θ)=1

cos (2 θ) + sin (θ) = 1

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Solution steps

cos (2 θ) + sin (θ) = 1

Subtract

1

from both sides

cos (2 θ) + sin (θ) - 1 = 0

Rewrite using trig identities

sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 0, sin (θ) = \frac{1}{2}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (x) - 2 cos (x) = 1

4 cos (x) - 1 = 0

sin (2 θ) = - \frac{1}{2}, 0 \leq θ < 2 π

sin (θ) = 1 - sin (θ)

2 sin (x) - 3 = 0, 0 \leq x \leq 2 π

What is the general solution for cos(2θ)+sin(θ)=1 ?
The general solution for cos(2θ)+sin(θ)=1 is θ=2pin,θ=pi+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin