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

The general solution for -2sin(θ)+cos(2θ)=1 is θ=(3pi)/2+2pin,θ=2pin,θ=pi+2pin

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-2sin(θ)+cos(2θ)=1

- 2 sin (θ) + cos (2 θ) = 1

θ = \frac{3 π}{2} + 2 πn, θ = 2 πn, θ = π + 2 πn

Solution steps

- 2 sin (θ) + cos (2 θ) = 1

Subtract

1

from both sides

- 2 sin (θ) + cos (2 θ) - 1 = 0

Rewrite using trig identities

- 2 sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = 0

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn, θ = 2 πn, θ = π + 2 πn

cos (2 x + 60) = sin (3 x)

csc (θ) = - \frac{3}{2}

4 cos (θ) = 0

\frac{sin ( 16 0 ^{\circ} )}{5} = \frac{sin ( c )}{4}

sin (x) + 2 = 0

What is the general solution for -2sin(θ)+cos(2θ)=1 ?
The general solution for -2sin(θ)+cos(2θ)=1 is θ=(3pi)/2+2pin,θ=2pin,θ=pi+2pin