Solve for x 2cos(x)^2=3sin(x)

$2 \cos^{2} (x) = 3 \sin (x)$

Step 1

Subtract $3 \sin (x)$ from both sides of the equation.

$2 \cos^{2} (x) - 3 \sin (x) = 0$

Step 2

Replace the $2 \cos^{2} (x)$ with $2 (1 - \sin^{2} (x))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$2 (1 - \sin^{2} (x)) - 3 \sin (x) = 0$

Step 3

Simplify each term.

Apply the distributive property.

$2 \cdot 1 + 2 (- \sin^{2} (x)) - 3 \sin (x) = 0$

Multiply $2$ by $1$ .

$2 + 2 (- \sin^{2} (x)) - 3 \sin (x) = 0$

Multiply $- 1$ by $2$ .

$2 - 2 \sin^{2} (x) - 3 \sin (x) = 0$

Step 4

Reorder the polynomial.

$- 2 \sin^{2} (x) - 3 \sin (x) + 2 = 0$

Step 5

Substitute $u$ for $\sin (x)$ .

$- 2 {(u)}^{2} - 3 u + 2 = 0$

Step 6

Factor the left side of the equation.

Factor $- 1$ out of $- 2 u^{2} - 3 u + 2$ .

Factor $- 1$ out of $- 2 u^{2}$ .

$- (2 u^{2}) - 3 u + 2 = 0$

Factor $- 1$ out of $- 3 u$ .

$- (2 u^{2}) - (3 u) + 2 = 0$

Rewrite $2$ as $- 1 (- 2)$ .

$- (2 u^{2}) - (3 u) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (2 u^{2}) - (3 u)$ .

$- (2 u^{2} + 3 u) - 1 \cdot - 2 = 0$

Factor $- 1$ out of $- (2 u^{2} + 3 u) - 1 (- 2)$ .

$- (2 u^{2} + 3 u - 2) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = 3$ .

Factor $3$ out of $3 u$ .

$- (2 u^{2} + 3 u - 2) = 0$

Rewrite $3$ as $- 1$ plus $4$

$- (2 u^{2} + (- 1 + 4) u - 2) = 0$

Apply the distributive property.

$- (2 u^{2} - 1 u + 4 u - 2) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- (2 u^{2} - 1 u + 4 u - 2) = 0$

Factor out the greatest common factor (GCF) from each group.

$- (u (2 u - 1) + 2 (2 u - 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

$- ((2 u - 1) (u + 2)) = 0$

Remove unnecessary parentheses.

$- (2 u - 1) (u + 2) = 0$

Step 7

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 u - 1 = 0$

$u + 2 = 0$

Step 8

Set $2 u - 1$ equal to $0$ and solve for $u$ .

Set $2 u - 1$ equal to $0$ .

$2 u - 1 = 0$

Solve $2 u - 1 = 0$ for $u$ .

Add $1$ to both sides of the equation.

$2 u = 1$

Divide each term in $2 u = 1$ by $2$ and simplify.

Divide each term in $2 u = 1$ by $2$ .

$\frac{2 u}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 u}{2} = \frac{1}{2}$

Divide $u$ by $1$ .

$u = \frac{1}{2}$

Step 9

Set $u + 2$ equal to $0$ and solve for $u$ .

Set $u + 2$ equal to $0$ .

$u + 2 = 0$

Subtract $2$ from both sides of the equation.

$u = - 2$

Step 10

The final solution is all the values that make $- (2 u - 1) (u + 2) = 0$ true.

$u = \frac{1}{2}, - 2$

Step 11

Substitute $\sin (x)$ for $u$ .

$\sin (x) = \frac{1}{2}, - 2$

Step 12

Set up each of the solutions to solve for $x$ .

$\sin (x) = \frac{1}{2}$

$\sin (x) = - 2$

Step 13

Solve for $x$ in $\sin (x) = \frac{1}{2}$ .

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (\frac{1}{2})$

Simplify the right side.

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - \frac{π}{6}$

Simplify $π - \frac{π}{6}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Find the period of $\sin (x)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 14

Solve for $x$ in $\sin (x) = - 2$ .

The range of sine is $- 1 \leq y \leq 1$ . Since $- 2$ does not fall in this range, there is no solution.

No solution

Step 15

List all of the solutions.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Solve for x 2cos(x)^2=3sin(x)

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