What is the general solution for solvefor θ,2sin(θ)=2cos(2θ) ?

The general solution for solvefor θ,2sin(θ)=2cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin

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solvefor θ,2sin(θ)=2cos(2θ)

solve for

θ, 2 sin (θ) = 2 cos (2 θ)

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

2 sin (θ) = 2 cos (2 θ)

Subtract

2 cos (2 θ)

from both sides

2 sin (θ) - 2 cos (2 θ) = 0

Rewrite using trig identities

- (1 - 2 sin^{2} (θ)) \cdot 2 + 2 sin (θ) = 0

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - 1

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

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

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

sec (θ) + 4 = 4 sec (θ)

sin (2 x) - 4 \cdot cos (x) = 0

5 tan (θ) + 3 = 0

sin (θ) = - \frac{4}{7}, π < θ < \frac{3 π}{2}, cos (2 θ)

tan (θ) = - \frac{2}{5}

What is the general solution for solvefor θ,2sin(θ)=2cos(2θ) ?
The general solution for solvefor θ,2sin(θ)=2cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin