Pracbice \( 1+\cos 2 \theta=-\frac{5}{6} \quad[180 ; 560] \) \( 1-\sin ^{2} \theta \) \( 2 \sin ^{2} \theta \) \( 5 \cos ^{2} \theta-1 \) \( 4 \cdot \sin ^{2} \theta-\cos ^{2} \theta \) solve \( =\frac{4 \sin x \cos x}{2 \sin 2-1} \) simplify \( =\frac{4 \sin 15 \cos 15}{2 \sin ^{2} 15-1}=\frac{2 \tan \theta \theta}{2} \) Prove \( =\frac{\sin ^{2} \theta}{\sin \theta}=4 \cos \theta-\frac{\cos 2 \theta+1}{\cos 2 \theta} \)

Let's break down the problem step by step, addressing each part of the problem systematically. ### Step 1: Solve the equation \( 1 + \cos 2\theta = -\frac{5}{6} \) We start with the equation: \[ 1 + \cos 2\theta = -\frac{5}{6} \] Subtracting 1 from both sides gives: \[ \cos 2\theta = -\frac{5}{6} - 1 = -\frac{5}{6} - \frac{6}{6} = -\frac{11}{6} \] However, since the cosine function cannot exceed 1 or be less than -1, we need to check if this equation is valid. The value \(-\frac{11}{6}\) is not possible for \(\cos 2\theta\). Therefore, there are no solutions for this equation. ### Step 2: Simplify \( 1 - \sin^2 \theta \) Using the Pythagorean identity: \[ 1 - \sin^2 \theta = \cos^2 \theta \] ### Step 3: Simplify \( 2 \sin^2 \theta \) This expression is already in its simplest form: \[ 2 \sin^2 \theta \] ### Step 4: Simplify \( 5 \cos^2 \theta - 1 \) Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \): \[ 5 \cos^2 \theta - 1 = 5(1 - \sin^2 \theta) - 1 = 5 - 5 \sin^2 \theta - 1 = 4 - 5 \sin^2 \theta \] ### Step 5: Simplify \( 4 \cdot \sin^2 \theta - \cos^2 \theta \) Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \): \[ 4 \sin^2 \theta - \cos^2 \theta = 4 \sin^2 \theta - (1 - \sin^2 \theta) = 4 \sin^2 \theta - 1 + \sin^2 \theta = 5 \sin^2 \theta - 1 \] ### Step 6: Solve \( = \frac{4 \sin x \cos x}{2 \sin 2 - 1} \) We need to clarify what \( x \) is in this context. Assuming \( x = \theta \): \[ \frac{4 \sin \theta \cos \theta}{2 \sin 2\theta - 1} \] Using the double angle identity \( \sin 2\theta = 2 \sin \theta \cos \theta \): \[ = \frac{4 \sin \theta \cos \theta}{2(2 \sin \theta \cos \theta) - 1} = \frac{4 \sin \theta \cos \theta}{4 \sin \theta \cos \theta - 1} \] ### Step 7: Simplify \( = \frac{4 \sin 15 \cos 15}{2 \sin^2 15 - 1} \) Using the double angle identity again: \[ = \frac{4 \sin 15 \cos 15}{2(1 - \cos 30)} = \frac{4 \sin 15 \cos 15}{2(1 - \frac{\sqrt{3}}{2})} = \frac{4 \sin 15 \cos 15}{2 \cdot \frac{2 - \sqrt{3}}{2}} = \frac{4 \sin 15 \cos 15}{2 - \sqrt{3}} \] ### Step 8: Prove \( = \frac{\sin^2 \theta}{\sin \theta} = 4 \cos \theta - \frac{\cos 2\theta + 1}{\cos 2\theta} \) Starting with the left side: \[ \frac{\sin^2 \theta}{\sin \theta} = \sin \theta \] Now, simplifying the right side: \[ 4 \cos \theta - \frac{\cos 2\theta + 1}{\cos 2\theta} = 4 \cos \theta - \left(1 + \frac{1}{\cos 2\theta}\right) \] Using the identity \( \cos 2\theta = 2\cos^2 \theta - 1 \): \[ = 4 \cos \theta - 1 - \frac{1}{2\cos^2 \theta - 1} \] This expression can be further simplified, but it requires specific values for \(\theta\) to prove equality. ### Conclusion The first equation has no solutions, while the other expressions have been simplified. The proof of the last equation requires specific values or further manipulation. If you have specific values for \(\theta\) or need further assistance, please let me know!