$$ \left. \begin{array} { l } { \sin ^ { 2 } x + 2 \sin x \cos x - 3 \cos ^ { 2 } x = 0 } \ { [ - 360 ; 360 ] } \end{array} ight. $$

Question

$$ \left. \begin{array} { l } { \sin ^ { 2 } x + 2 \sin x \cos x - 3 \cos ^ { 2 } x = 0 } \ { [ - 360 ; 360 ] } \end{array} ight. $$

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\sin^{2}\left(x\right)+2\sin\left(x\right)\cos\left(x\right)-3\cos^{2}\left(x\right)=0$$
- step1: Factor the expression:
  $$\left(\sin\left(x\right)+3\cos\left(x\right)\right)\left(\sin\left(x\right)-\cos\left(x\right)\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&\sin\left(x\right)+3\cos\left(x\right)=0\&\sin\left(x\right)-\cos\left(x\right)=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=\operatorname{arccot}\left(-\frac{1}{3}\right)+k\pi ,k \in \mathbb{Z}\&x=\frac{\pi }{4}+k\pi ,k \in \mathbb{Z}\end{align}$$
- step4: Find the union:
  $$x=\left\{ \begin{array}{l}\frac{\pi }{4}+k\pi \\operatorname{arccot}\left(-\frac{1}{3}\right)+k\pi \end{array}\right.,k \in \mathbb{Z}$$
Solve the equation $$ x = \frac{\pi}{4} + k\pi $$.
Solve the equation by following steps:
- step0: Solve for $$k$$:
  $$x=\frac{\pi }{4}+k\pi $$
- step1: Reorder the terms:
  $$x=\frac{\pi }{4}+\pi k$$
- step2: Swap the sides:
  $$\frac{\pi }{4}+\pi k=x$$
- step3: Move the constant to the right side:
  $$\pi k=x-\frac{\pi }{4}$$
- step4: Subtract the terms:
  $$\pi k=\frac{4x-\pi }{4}$$
- step5: Multiply by the reciprocal:
  $$\pi k\times \frac{1}{\pi }=\frac{4x-\pi }{4}\times \frac{1}{\pi }$$
- step6: Multiply:
  $$k=\frac{4x-\pi }{4\pi }$$
Solve the equation $$ x = \operatorname{arccot}(-\frac{1}{3}) + k\pi $$.
Solve the equation by following steps:
- step0: Solve for $$k$$:
  $$x=\operatorname{arccot}\left(-\frac{1}{3}\right)+k\pi $$
- step1: Reorder the terms:
  $$x=\operatorname{arccot}\left(-\frac{1}{3}\right)+\pi k$$
- step2: Swap the sides:
  $$\operatorname{arccot}\left(-\frac{1}{3}\right)+\pi k=x$$
- step3: Move the constant to the right side:
  $$\pi k=x-\operatorname{arccot}\left(-\frac{1}{3}\right)$$
- step4: Divide both sides:
  $$\frac{\pi k}{\pi }=\frac{x-\operatorname{arccot}\left(-\frac{1}{3}\right)}{\pi }$$
- step5: Divide the numbers:
  $$k=\frac{x-\operatorname{arccot}\left(-\frac{1}{3}\right)}{\pi }$$
To solve the equation

$$
\sin^2 x + 2 \sin x \cos x - 3 \cos^2 x = 0
$$

within the interval $$[-360, 360]$$, we can start by rewriting the equation in terms of a single trigonometric function.

### Step 1: Rewrite the equation

Using the identity $$\sin^2 x = 1 - \cos^2 x$$, we can express everything in terms of $$\cos x$$:

$$
(1 - \cos^2 x) + 2 \sin x \cos x - 3 \cos^2 x = 0
$$

This simplifies to:

$$
1 - 4 \cos^2 x + 2 \sin x \cos x = 0
$$

### Step 2: Substitute $$\sin x$$

Using the identity $$\sin x = \sqrt{1 - \cos^2 x}$$, we can substitute $$\sin x$$ into the equation. However, it is more straightforward to solve the original equation directly.

### Step 3: Factor the equation

The equation can be factored as:

$$
(\sin x - 3 \cos x)(\sin x + \cos x) = 0
$$

This gives us two separate equations to solve:

1. $$\sin x - 3 \cos x = 0$$
2. $$\sin x + \cos x = 0$$

### Step 4: Solve each equation

#### For $$\sin x - 3 \cos x = 0$$:

This can be rewritten as:

$$
\sin x = 3 \cos x
$$

Dividing both sides by $$\cos x$$ (assuming $$\cos x \neq 0$$) gives:

$$
\tan x = 3
$$

The general solution for this is:

$$
x = \arctan(3) + k\pi, \quad k \in \mathbb{Z}
$$

#### For $$\sin x + \cos x = 0$$:

This can be rewritten as:

$$
\sin x = -\cos x
$$

Dividing both sides by $$\cos x$$ (assuming $$\cos x \neq 0$$) gives:

$$
\tan x = -1
$$

The general solution for this is:

$$
x = \frac{3\pi}{4} + k\pi, \quad k \in \mathbb{Z}
$$

### Step 5: Find specific solutions in the interval $$[-360, 360]$$

1. For $$x = \arctan(3) + k\pi$$:
   - Calculate $$\arctan(3)$$ and find values for $$k$$ such that $$x$$ is within $$[-360, 360]$$.

2. For $$x = \frac{3\pi}{4} + k\pi$$:
   - Substitute values for $$k$$ to find solutions within the interval.

### Final Solutions

The solutions can be expressed as:

$$
x = \left\{ \frac{\pi}{4} + k\pi, \arccot(-\frac{1}{3}) + k\pi \right\}, \quad k \in \mathbb{Z}
$$

Now, we can calculate specific values for $$k$$ to find the solutions within the specified interval.

Let's calculate the specific values for $$k$$ for both cases.

1. For $$x = \frac{\pi}{4} + k\pi$$:
   - $$k = -2$$ gives $$x = \frac{\pi}{4} - 2\pi = \frac{\pi}{4} - \frac{8\pi}{4} = -\frac{7\pi}{4}$$
   - $$k = -1$$ gives $$x = \frac{\pi}{4} - \pi = \frac{\pi}{4} - \frac{4\pi}{4} = -\frac{3\pi}{4}$$
   - $$k = 0$$ gives $$x = \frac{\pi}{4}$$
   - $$k = 1$$ gives $$x = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$$
   - $$k = 2$$ gives $$x = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4}$$ (out of range)

2. For $$x = \arccot(-\frac{1}{3}) + k\pi$$:
   - Calculate $$\arccot(-\frac{1}{3})$$ and find values for $$k$$.

Now, let's calculate $$\arccot(-\frac{1}{3})$$ and find the specific values.

### Conclusion

The final solutions in the interval $$[-360, 360]$$ will be:

- From $$\tan x = 3$$: $$x = \arctan(3) + k\pi$$
- From $$\tan x = -1$$: $$x = \frac{3\pi}{4} + k\pi$$

Calculating these will yield the specific angles.
The solutions to the equation $$ \sin^2 x + 2 \sin x \cos x - 3 \cos^2 x = 0 $$ within the interval $$[-360^\circ, 360^\circ]$$ are: 1. $$ x = \arctan(3) + k\pi $$, where $$ k $$ is an integer. 2. $$ x = \frac{3\pi}{4} + k\pi $$, where $$ k $$ is an integer. Specific solutions within the interval are: - $$ x = -\frac{7\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}, \frac{5\pi}{4} $$. These are the values of $$ x $$ that satisfy the equation within the given range.

\( \left. \begin{array} { l } { \sin ^ { 2 } x + 2 \sin x \cos x - 3 \cos ^ { 2 } x = 0 } \\ { [ - 360 ; 360 ] } \end{array} \right. \)

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