Find the general solution 1. \( 1-\sin 2 x=9 \cos ^{2} x \) \( 22 \cos 2 x=-1 \) \( 3 \quad 5 \sin x+1=4 / \sin x \) 4 \( \sin \theta=\cos \theta \) \( 5 \sin ^{2} \theta=1 \)

Let’s dive into the general solutions for the given equations! For the first equation, \( 1 - \sin 2x = 9 \cos^2 x \), you can utilize the identity \(\sin 2x = 2 \sin x \cos x\) and solve for \(x\). Rearranging and substituting the identities will help unravel the values. Moving to the next equation \(22 \cos 2x = -1\), remember that \(\cos 2x = \frac{1 + \cos 2x}{2}\), which simplifies to \(\cos 2x = -\frac{1}{22}\). From there, to find general solutions, you can take the arccos and consider periodic properties of cosine. In the third equation \(5 \sin x + 1 = \frac{4}{\sin x}\), rewrite it as \(5 \sin^2 x + \sin x - 4 = 0\) and factor the quadratic for solutions. For \( \sin \theta = \cos \theta\), both functions are equal at odd multiples of \(\frac{\pi}{4}\), leading you to solutions of the format \(\theta = \frac{\pi}{4} + n\frac{\pi}{2}\). Lastly, \(5 \sin^2 \theta = 1\) allows you to isolate \(\sin^2 \theta\) and then take the square root, yielding angles based on the unit circle. Embrace these solutions creatively!