What is the general solution for 2+2sin(θ)=4cos^2(θ) ?

The general solution for 2+2sin(θ)=4cos^2(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin

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2+2sin(θ)=4cos^2(θ)

2 + 2 sin (θ) = 4 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 + 2 sin (θ) = 4 cos^{2} (θ)

Subtract

4 cos^{2} (θ)

from both sides

2 + 2 sin (θ) - 4 cos^{2} (θ) = 0

Rewrite using trig identities

- 2 + 2 sin (θ) + 4 sin^{2} (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

9 tan (x) sin (x) - 5 sin (x) = 0

x, y = arctan (x)

7 cos (x) + 7 sin (x) tan (x) = 14

cot (θ) + 5 csc (θ) = 6

3 csc^{2} (x) + 2 csc (x) = 0

What is the general solution for 2+2sin(θ)=4cos^2(θ) ?
The general solution for 2+2sin(θ)=4cos^2(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin