1. Solve \( \cos ^{2} \theta-\sin \theta-1=0 \) Solve \( 2 \cos 2 \theta+\cos \theta+2=0 \) a) Solve \( 10 \cos ^{2} \theta-5 \sin 2 \theta+2=0 \). b) Use the solution to a) above to obtain values in the range \( \left[-270^{\circ} ; 180^{\circ}\right] \) whic satisfy the equation.

\( \textbf{Problem 1.} \) Solve \[ \cos^2 \theta - \sin \theta - 1 = 0. \] 1. Express \(\cos^2 \theta\) in terms of \(\sin \theta\) using the identity \[ \cos^2\theta = 1 - \sin^2\theta. \] Then \[ 1 - \sin^2\theta - \sin\theta - 1 = 0 \quad \Longrightarrow \quad -\sin^2\theta - \sin\theta = 0. \] 2. Multiply by \(-1\) to obtain \[ \sin^2\theta + \sin\theta = 0. \] 3. Factor the left‐side: \[ \sin\theta(\sin\theta+1)=0. \] 4. So the solutions come from \[ \sin\theta=0 \quad \text{or} \quad \sin\theta=-1. \] 5. Hence, in general: - If \(\sin\theta = 0\), then \[ \theta = k\pi\quad (\text{or } \theta = 0^\circ, 180^\circ, 360^\circ,\dots) \] for any integer \(k\). - If \(\sin\theta = -1\), then \[ \theta = \frac{3\pi}{2}+2\pi k\quad (\text{or } \theta = 270^\circ+360^\circ k) \] for any integer \(k\). --- \( \textbf{Problem 2.} \) Solve \[ 2\cos2\theta + \cos\theta + 2 = 0. \] 1. Write \(\cos2\theta\) in terms of \(\cos^2\theta\) by using \[ \cos2\theta = 2\cos^2\theta - 1. \] 2. Substitute into the equation: \[ 2\bigl(2\cos^2\theta-1\bigr)+\cos\theta+2 = 0, \] which becomes \[ 4\cos^2\theta - 2 + \cos\theta + 2 = 0. \] 3. Simplify: \[ 4\cos^2\theta + \cos\theta = 0. \] 4. Factor out \(\cos\theta\): \[ \cos\theta\bigl(4\cos\theta+1\bigr)=0. \] 5. Therefore, the solutions are given by - \(\cos\theta=0\), so \[ \theta = \frac{\pi}{2} + \pi k\quad (\text{or }90^\circ+180^\circ k) \] for any integer \(k\); - \(4\cos\theta+1=0\), i.e., \[ \cos\theta=-\frac{1}{4}. \] Thus, \[ \theta = \arccos\Bigl(-\frac{1}{4}\Bigr)+2\pi k\quad \text{or}\quad \theta = -\arccos\Bigl(-\frac{1}{4}\Bigr)+2\pi k, \] for any integer \(k\). --- \( \textbf{Problem 3. (a)} \) Solve \[ 10\cos^2\theta - 5\sin2\theta + 2 = 0. \] 1. Write \(\cos^2\theta\) in terms of \(\cos2\theta\) by means of the double‐angle formula: \[ \cos2\theta = 2\cos^2\theta - 1 \quad \Longrightarrow \quad \cos^2\theta = \frac{1+\cos2\theta}{2}. \] 2. Substitute into the equation: \[ 10\left(\frac{1+\cos2\theta}{2}\right)-5\sin2\theta+2=0. \] That is, \[ 5(1+\cos2\theta)-5\sin2\theta+2=0. \] 3. Simplify: \[ 5+5\cos2\theta-5\sin2\theta+2=0 \quad \Longrightarrow \quad 5\cos2\theta-5\sin2\