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

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

Subtract $2 \cos^{2} (t)$ from both sides of the equation.

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

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

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 2$ by $1$ .

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

Multiply $- 1$ by $- 2$ .

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

Reorder the polynomial.

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

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

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

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, $2 u - 1$ .

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

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

$2 u - 1 = 0$

$u + 2 = 0$

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 $1$ .

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

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.

$u = - 2$

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

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

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

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

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

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

$\sin (t) = - 2$

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

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

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

Simplify the right side.

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

$t = \frac{π}{6}$

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.

$t = π - \frac{π}{6}$

Simplify $π - \frac{π}{6}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

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

Combine the numerators over the common denominator.

$t = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$t = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$t = \frac{5 π}{6}$

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

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 $\sin (t)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

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

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

No solution

List all of the solutions.

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

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

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