How in the h**l do yoy solve sin^4 2 theta - 2 sin^2 2 theta = -1

(t=theta)

Set sin(2*t)^2=x

solve the polynomial x^2 - 2x = -1

then all you have to do is solve sin(2t)=sqrt(x1) and sin(2t)=sqrt(x2) where x1, and x2 are the solutions for the polynomial.

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sin^4 2 theta - 2 sin^2 2 theta = -1

\(sin^4 2 \theta - 2 sin^2 2 \theta = -1\\ (sin^2 2\theta)^2 - 2 (sin^2 2\theta )= -1\\ (sin^2 2\theta)^2 - 2 (sin^2 2\theta )+1=0\\ \text{let }x=sin^22\theta\\ x^2 - 2x+1=0\\ (x-1)(x-1)=0\\ x=+1\\ so\\ sin^22\theta=1\\ sin2\theta=\pm1\\ 2\theta=\frac{\pi}{2},\;\;\frac{3\pi}{2},.......\\ 2\theta=\frac{\pi}{2}+n\pi \qquad n\in Z\;\;(n \;is\;an\; integer)\\ \theta=\frac{\pi}{4}+\frac{n\pi}{2} \qquad n\in Z\\ \theta=\frac{\pi(1+2n)}{4} \qquad n\in Z\\ \)

check:

\(sin^4 2 \theta - 2 sin^2 2 \theta = -1\\ LHS\\ =\left(sin \frac{\pi(1+2n)}{2}\right)^4 - 2 \left(sin \frac{\pi(1+2n)}{2}\right)^2\\ =\left(\pm1\right)^4 - 2 \left(\pm1\right)^2\\ =1 - 2 *1\\ =-1\\ =RHS\)

check: