\begin{tabular}{ll}10 & \( 1-5 \sin ^{2} x=3 \sin x+1 \) and \( x \in\left[-180^{\circ} ; 180^{\circ}\right] \) \\ 11 & \( 6 \sin ^{2} x+7 \sin x+2=0 \) and \( x \in\left[-360^{\circ} ; 360^{\circ}\right] \) \\ 12 & \( 3 \cos ^{2} x+\mathrm{R} 3 \sin x \cos x=0 ; \) find GS \\ 13 & \( 8 \sin x \cos x+4 \cos x+6 \sin x+3=0 ; \) find GS \\ 14 & \( 4 \cos ^{2} x-\sin ^{2} x=3 \cos x+1 \) and \( x \in\left[-180^{\circ} ; 180^{\circ}\right] \) \\ 15 & \( 3 \sin ^{2} x+2 \sin x \cdot \cos x=4 \cos x+6 \sin x \) and \( x \in\left[0^{\circ} ; 36\right. \) \\ \hline\end{tabular}

Solve the equation by following steps: - step0: Solve for \(x\): \(8\sin\left(x\right)\cos\left(x\right)+4\cos\left(x\right)+6\sin\left(x\right)+3=0\) - step1: Factor the expression: \(\left(2\sin\left(x\right)+1\right)\left(4\cos\left(x\right)+3\right)=0\) - step2: Separate into possible cases: \(\begin{align}&2\sin\left(x\right)+1=0\\&4\cos\left(x\right)+3=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=\left\{ \begin{array}{l}\frac{7\pi }{6}+2k\pi \\\frac{11\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\\&x=\left\{ \begin{array}{l}-\arccos\left(-\frac{3}{4}\right)+2k\pi \\\arccos\left(-\frac{3}{4}\right)+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\) - step4: Find the union: \(x=\left\{ \begin{array}{l}-\arccos\left(-\frac{3}{4}\right)+2k\pi \\\arccos\left(-\frac{3}{4}\right)+2k\pi \\\frac{7\pi }{6}+2k\pi \\\frac{11\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) Let's solve the equations step by step based on the provided problems. ### Problem 10: \( 1 - 5 \sin^2 x = 3 \sin x + 1 \) 1. Rearranging the equation: \[ 1 - 5 \sin^2 x - 3 \sin x - 1 = 0 \implies -5 \sin^2 x - 3 \sin x = 0 \] This simplifies to: \[ 5 \sin^2 x + 3 \sin x = 0 \] Factoring out \(\sin x\): \[ \sin x (5 \sin x + 3) = 0 \] This gives us two cases: - \(\sin x = 0\) - \(5 \sin x + 3 = 0 \implies \sin x = -\frac{3}{5}\) 2. Solving \(\sin x = 0\): - \(x = 0^\circ, 180^\circ\) 3. Solving \(\sin x = -\frac{3}{5}\): - \(x = \arcsin(-\frac{3}{5})\) and its reference angle in the fourth quadrant. ### Problem 11: \( 6 \sin^2 x + 7 \sin x + 2 = 0 \) This is a quadratic equation in terms of \(\sin x\). We can use the quadratic formula: \[ \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 6\), \(b = 7\), and \(c = 2\). Calculating the discriminant: \[ D = 7^2 - 4 \cdot 6 \cdot 2 = 49 - 48 = 1 \] Thus, \[ \sin x = \frac{-7 \pm 1}{12} \implies \sin x = -\frac{6}{12} = -\frac{1}{2} \quad \text{or} \quad \sin x = -\frac{8}{12} = -\frac{2}{3} \] ### Problem 12: \( 3 \cos^2 x + R \cdot 3 \sin x \cos x = 0 \) This equation can be factored as: \[ 3 \cos x ( \cos x + R \sin x) = 0 \] This gives us: - \(\cos x = 0\) which leads to \(x = 90^\circ, -90^\circ\) - \(\cos x + R \sin x = 0\) which can be solved for specific values of \(R\). ### Problem 13: \( 8 \sin x \cos x + 4 \cos x + 6 \sin x + 3 = 0 \) Factoring out \(\cos x\): \[ \cos x (8 \sin x + 4) + 6 \sin x + 3 = 0 \] This can be solved for specific values of \(\sin x\) and \(\cos x\). ### Problem 14: \( 4 \cos^2 x - \sin^2 x = 3 \cos x + 1 \) Rearranging gives: \[ 4 \cos^2 x - \sin^2 x - 3 \cos x - 1 = 0 \] Using \(\sin^2 x = 1 - \cos^2 x\) allows us to express everything in terms of \(\cos x\). ### Problem 15: \( 3 \sin^2 x + 2 \sin x \cos x = 4 \cos x + 6 \sin x \) Rearranging gives: \[ 3 \sin^2 x + 2 \sin x \cos x - 4 \cos x - 6 \sin x = 0 \] This can be solved using substitution or factoring. ### Summary of Results - For Problem 10, we have solutions for \(\sin x = 0\) and \(\sin x = -\frac{3}{5}\). - For Problem 11, we have \(\sin x = -\frac{1}{2}\) and \(\sin x = -\frac{2}{3}\). - Problems 12, 13, 14, and 15 require specific values or further simplification based on the context of \(R\) or additional constraints. If you need further calculations or specific values for \(R\) or any other details, please let me know!