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

The general solution for -sin(θ)+cos(2θ)=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin,θ=pi+2pin

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-sin(θ)+cos(2θ)=1

- sin (θ) + cos (2 θ) = 1

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn, θ = π + 2 πn

Solution steps

- sin (θ) + cos (2 θ) = 1

Subtract

1

from both sides

- sin (θ) + cos (2 θ) - 1 = 0

Rewrite using trig identities

- sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - \frac{1}{2}, sin (θ) = 0

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn, θ = π + 2 πn

arctan (2 x - 8) = - 1

- 2 cos (2 θ) - 5 = - 4 sin (θ) - 8

3 cos^{2} (θ) - cos (θ) = 0

10 sin^{2} (x) + 11 sin (x) - 8 = 0

2 cos^{2} (x) + 7 cos (x) - 4 = 0

What is the general solution for -sin(θ)+cos(2θ)=1 ?
The general solution for -sin(θ)+cos(2θ)=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin,θ=pi+2pin