Solve for x square root of 2sin(x)^2+cos(x)=0

Solution:

Step 1:

Utilize the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$ to express $\sin^2(x)$ as $1 - \cos^2(x)$.

$$\sqrt{2}(1 - \cos^2(x)) + \cos(x) = 0$$

Step 2:

Begin simplification of the equation.

Step 2.1:

Distribute $\sqrt{2}$ across the terms inside the parentheses.

$$\sqrt{2} \cdot 1 - \sqrt{2}\cos^2(x) + \cos(x) = 0$$

Step 2.2:

Perform the multiplication of $\sqrt{2}$ by 1.

$$\sqrt{2} - \sqrt{2}\cos^2(x) + \cos(x) = 0$$

Step 3:

Rearrange the terms to form a polynomial.

$$-\sqrt{2}\cos^2(x) + \cos(x) + \sqrt{2} = 0$$

Step 4:

Introduce a substitution, let $u = \cos(x)$.

$$-\sqrt{2}u^2 + u + \sqrt{2} = 0$$

Step 5:

Apply the quadratic formula to find $u$.

$$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Step 6:

Insert the coefficients $a = -\sqrt{2}$, $b = 1$, and $c = \sqrt{2}$ into the formula.

$$u = \frac{-1 \pm \sqrt{1^2 - 4(-\sqrt{2})(\sqrt{2})}}{2(-\sqrt{2})}$$

Step 7:

Proceed with simplification.

Step 7.1:

Simplify the numerator.

Step 7.1.1:

Recognize that $1$ raised to any power is still $1$.

$$u = \frac{-1 \pm \sqrt{1 + 4\sqrt{2}\sqrt{2}}}{2(-\sqrt{2})}$$

Step 7.1.2:

Calculate the product of $-4$ and $-\sqrt{2}\sqrt{2}$.

$$u = \frac{-1 \pm \sqrt{1 + 4(\sqrt{2})^2}}{2(-\sqrt{2})}$$

Step 7.1.3:

Combine the square roots.

$$u = \frac{-1 \pm \sqrt{1 + 4 \cdot 2}}{2(-\sqrt{2})}$$

Step 7.1.4:

Add the numbers inside the square root.

$$u = \frac{-1 \pm \sqrt{9}}{2(-\sqrt{2})}$$

Step 7.1.5:

Recognize that $\sqrt{9} = 3$.

$$u = \frac{-1 \pm 3}{2(-\sqrt{2})}$$

Step 7.2:

Multiply the numerator by $-1$.

$$u = \frac{1 \pm 3}{2\sqrt{2}}$$

Step 7.3:

Rationalize the denominator by multiplying by $\frac{\sqrt{2}}{\sqrt{2}}$.

$$u = \frac{(1 \pm 3)\sqrt{2}}{2\sqrt{2}\sqrt{2}}$$

Step 7.4:

Simplify the denominator.

$$u = \frac{(1 \pm 3)\sqrt{2}}{4}$$

Step 8:

Combine the solutions for $u$.

$$u = \sqrt{2}, -\frac{\sqrt{2}}{2}$$

Step 9:

Replace $u$ with $\cos(x)$.

$$\cos(x) = \sqrt{2}, -\frac{\sqrt{2}}{2}$$

Step 10:

Solve for $x$ for each value of $\cos(x)$.

Step 10.1:

For $\cos(x) = \sqrt{2}$, there is no solution as $\sqrt{2}$ is outside the range of the cosine function.

Step 10.2:

For $\cos(x) = -\frac{\sqrt{2}}{2}$, find the angle whose cosine is $-\frac{\sqrt{2}}{2}$.

$$x = \arccos\left(-\frac{\sqrt{2}}{2}\right)$$

Step 11:

Determine the exact values for $x$.

$$x = \frac{3\pi}{4}, \frac{5\pi}{4}$$

Step 12:

Include the periodic nature of the cosine function to list all solutions.

$$x = \frac{3\pi}{4} + 2\pi n, \frac{5\pi}{4} + 2\pi n$$ where $n$ is any integer.

Step 13:

The final solutions are:

$$x = \frac{3\pi}{4} + 2\pi n, \frac{5\pi}{4} + 2\pi n$$ for any integer $n$.

Knowledge Notes:

The problem involves solving a trigonometric equation that includes a square root and both sine and cosine functions. The key steps in solving this problem include:

1. Pythagorean Identity: This identity states that $\sin^2(x) + \cos^2(x) = 1$. It is used to express $\sin^2(x)$ in terms of $\cos^2(x)$, which simplifies the equation to a quadratic form.

2. Quadratic Equations: The transformed equation resembles a quadratic equation in terms of $\cos(x)$, which can be solved using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

3. Rationalizing the Denominator: When a square root appears in the denominator, it is often rationalized to simplify the expression.

4. Inverse Trigonometric Functions: To solve for the angle $x$, the inverse cosine function $\arccos$ is used, which gives the angle whose cosine is the given value.

5. Trigonometric Function Range: The cosine function has a range from $-1$ to $1$. Any value outside this range does not have an angle whose cosine equals that value.

6. Periodicity of Trigonometric Functions: The cosine function is periodic with a period of $2\pi$. Therefore, once the principal solutions are found, all other solutions can be found by adding integer multiples of $2\pi$.

7. Exact Values: Some trigonometric equations have exact values that can be determined without a calculator, such as $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$.

By applying these concepts, the problem is reduced to a more manageable form, and the solutions are found within the domain of the cosine function.