What is the general solution for 1=4sin^2(θ) ?

The general solution for 1=4sin^2(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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1=4sin^2(θ)

1 = 4 sin^{2} (θ)

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solution steps

1 = 4 sin^{2} (θ)

Switch sides

4 sin^{2} (θ) = 1

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{2}

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

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ, 1 = sin (θ)

tan (2 x) = 2

\frac{sin ( b )}{13.1} = \frac{sin ( 112. 1 ^{\circ} )}{24.1}

tan (α) = \frac{0.006}{π \cdot 0.027}

sec^{2} (θ) tan (θ) = 2 tan (θ)

What is the general solution for 1=4sin^2(θ) ?
The general solution for 1=4sin^2(θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(7pi)/6+2pin,θ=(11pi)/6+2pin