\begin{tabular}{ll}10 & $$ 1-5 \sin ^{2} x=3 \sin x+1 $$ and $$ x \in\left[-180^{\circ} ; 180^{\circ} ight] $$ \ 11 & $$ 6 \sin ^{2} x+7 \sin x+2=0 $$ and $$ x \in\left[-360^{\circ} ; 360^{\circ} ight] $$ \ 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 $$ x \in\left[-180^{\circ} ; 180^{\circ} ight] $$ \ 15 & $$ 3 \sin ^{2} x+2 \sin x \cdot \cos x=4 \cos x+6 \sin x $$ and $$ x \in\left[0^{\circ} ; 36 ight. $$ \ \hline\end{tabular}

Question

\begin{tabular}{ll}10 & $$ 1-5 \sin ^{2} x=3 \sin x+1 $$ and $$ x \in\left[-180^{\circ} ; 180^{\circ}ight] $$ \ 11 & $$ 6 \sin ^{2} x+7 \sin x+2=0 $$ and $$ x \in\left[-360^{\circ} ; 360^{\circ}ight] $$ \ 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 $$ x \in\left[-180^{\circ} ; 180^{\circ}ight] $$ \ 15 & $$ 3 \sin ^{2} x+2 \sin x \cdot \cos x=4 \cos x+6 \sin x $$ and $$ x \in\left[0^{\circ} ; 36ight. $$ \ \hline\end{tabular}

UpStudy AI · Accepted Answer

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!
Solutions for the equations are as follows: 1. **Problem 10**: $$x = 0^\circ, 180^\circ$$, and $$x = \arcsin(-\frac{3}{5})$$ in the fourth quadrant. 2. **Problem 11**: $$x = 210^\circ, 330^\circ$$, and $$x = 240^\circ, 300^\circ$$. 3. **Problem 12**: $$x = 90^\circ, -90^\circ$$, and $$x = \arctan(-\frac{R}{3})$$ in the appropriate quadrants. 4. **Problem 13**: $$x = 45^\circ, 135^\circ$$, and $$x = 225^\circ, 315^\circ$$. 5. **Problem 14**: $$x = 60^\circ, -60^\circ$$, and $$x = 120^\circ, -120^\circ$$. 6. **Problem 15**: $$x = 30^\circ, 150^\circ$$, and $$x = 210^\circ, 330^\circ$$. These are the general solutions within the specified intervals.

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