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

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

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2sin(2θ)=2cos(θ)

2 sin (2 θ) = 2 cos (θ)

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

Solution steps

2 sin (2 θ) = 2 cos (θ)

Subtract

2 cos (θ)

from both sides

2 sin (2 θ) - 2 cos (θ) = 0

Rewrite using trig identities

- 2 cos (θ) + 4 cos (θ) sin (θ) = 0

Factor

- 2 cos (θ) + 4 cos (θ) sin (θ) : 2 cos (θ) (2 sin (θ) - 1)

2 cos (θ) (2 sin (θ) - 1) = 0

Solving each part separately

cos (θ) = 0 or 2 sin (θ) - 1 = 0

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

2 sin (θ) - 1 = 0 : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

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

1 + 3 tan (\frac{2 x - π}{3}) = 0

tan (3 x) = tan (π x)

sin^{2} (2 x) = 0.5

sin (x) = - \frac{1}{2} 2

(6 cos (x) + 5 sin (x))^{2} + 11 sin^{2} (x) = 66

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