What is the general solution for 2-sin(2θ)=4sin(2θ) ?

The general solution for 2-sin(2θ)=4sin(2θ) is θ=(0.41151…)/2+pin,θ= pi/2-(0.41151…)/2+pin

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2-sin(2θ)=4sin(2θ)

2 - sin (2 θ) = 4 sin (2 θ)

θ = \frac{0.41151 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.41151 \dots}{2} + πn

Solution steps

2 - sin (2 θ) = 4 sin (2 θ)

Solve by substitution

sin (2 θ) = \frac{2}{5}

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

Combine all the solutions

θ = \frac{arcsin ( \frac{2}{5} )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( \frac{2}{5} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.41151 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.41151 \dots}{2} + πn

2 sin (θ) cos (θ) + sin (θ) = 0

csc (3 x) - 8 = 0

sin^{2} (x) = tan (x)

sin (x) \cdot cos (x) = \frac{1}{2}

cos (x) sin (x) - cos (x) = 0

What is the general solution for 2-sin(2θ)=4sin(2θ) ?
The general solution for 2-sin(2θ)=4sin(2θ) is θ=(0.41151…)/2+pin,θ= pi/2-(0.41151…)/2+pin