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

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

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2sin(θ)+cos(2θ)=1

2 sin (θ) + cos (2 θ) = 1

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{2} + 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) = 0, sin (θ) = 1

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

Combine all the solutions

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

sin (π t) = 0

8 sin^{3} (x) = 8 sin (x)

sin^{2} (x) + cos^{2} (x) = sin (x)

cos (θ) + 2 sec (θ) = - 3

3 tan^{2} (θ) + 7 tan (θ) - 2 = 4

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