What is the general solution for 4sin^2(2θ)+6=9 ?

The general solution for 4sin^2(2θ)+6=9 is θ= pi/6+pin,θ= pi/3+pin,θ=(2pi)/3+pin,θ=(5pi)/6+pin

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4sin^2(2θ)+6=9

4 sin^{2} (2 θ) + 6 = 9

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

Solution steps

4 sin^{2} (2 θ) + 6 = 9

Solve by substitution

sin (2 θ) = \frac{3}{2}, sin (2 θ) = - \frac{3}{2}

sin (2 θ) = \frac{3}{2} : θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

sin (2 θ) = - \frac{3}{2} : θ = \frac{2 π}{3} + πn, θ = \frac{5 π}{6} + πn

Combine all the solutions

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

tan (a) = \frac{1}{3}

tan (x) = 0.649

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} = tan (x)

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

tan (θ) = \frac{1}{2}

What is the general solution for 4sin^2(2θ)+6=9 ?
The general solution for 4sin^2(2θ)+6=9 is θ= pi/6+pin,θ= pi/3+pin,θ=(2pi)/3+pin,θ=(5pi)/6+pin