What is the general solution for tan(2θ)+tan(2θ)sin(2θ)=0 ?

The general solution for tan(2θ)+tan(2θ)sin(2θ)=0 is θ=(pin)/2

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tan(2θ)+tan(2θ)sin(2θ)=0

tan (2 θ) + tan (2 θ) sin (2 θ) = 0

θ = \frac{πn}{2}

Solution steps

tan (2 θ) + tan (2 θ) sin (2 θ) = 0

Factor

tan (2 θ) + tan (2 θ) sin (2 θ) : tan (2 θ) (sin (2 θ) + 1)

tan (2 θ) (sin (2 θ) + 1) = 0

Solving each part separately

tan (2 θ) = 0 or sin (2 θ) + 1 = 0

tan (2 θ) = 0 : θ = \frac{πn}{2}

sin (2 θ) + 1 = 0 : θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = \frac{πn}{2}, θ = \frac{3 π}{4} + πn

Since the equation is undefined for:

\frac{3 π}{4} + πn

θ = \frac{πn}{2}

3 cot (\frac{1}{4} x) = 3

4 sin (x) - cos (2 x) = 1

2 sin^{2} (t) - 3 sin (t) + 1 = 0

5 sin (x) = 0

5 sin (x) = 4

What is the general solution for tan(2θ)+tan(2θ)sin(2θ)=0 ?
The general solution for tan(2θ)+tan(2θ)sin(2θ)=0 is θ=(pin)/2