Solve for θ in Degrees 2sin(theta)^2-3sin(theta)+1=0

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Trigonometry

$2 \sin^{2} (θ) - 3 \sin (θ) + 1 = 0$

Factor the left side of the equation.

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Let $u = \sin (θ)$ . Substitute $u$ for all occurrences of $\sin (θ)$ .

$2 u^{2} - 3 u + 1 = 0$

Factor by grouping.

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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 $a \cdot c = 2 \cdot 1 = 2$ and whose sum is $b = - 3$ .

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

$2 u^{2} - 3 u + 1 = 0$

Rewrite $- 3$ as $- 1$ plus $- 2$

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

Apply the distributive property.

$2 u^{2} - 1 u - 2 u + 1 = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

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

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

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

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

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

Replace all occurrences of $u$ with $\sin (θ)$ .

$(2 \sin (θ) - 1) (\sin (θ) - 1) = 0$

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

$2 \sin (θ) - 1 = 0$

$\sin (θ) - 1 = 0$

Set $2 \sin (θ) - 1$ equal to $0$ and solve for $θ$ .

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Set $2 \sin (θ) - 1$ equal to $0$ .

$2 \sin (θ) - 1 = 0$

Solve $2 \sin (θ) - 1 = 0$ for $θ$ .

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Add $1$ to both sides of the equation.

$2 \sin (θ) = 1$

Divide each term in $2 \sin (θ) = 1$ by $2$ and simplify.

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Divide each term in $2 \sin (θ) = 1$ by $2$ .

$\frac{2 \sin (θ)}{2} = \frac{1}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \sin (θ)}{2} = \frac{1}{2}$

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{1}{2}$

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

$θ = \arcsin (\frac{1}{2})$

Simplify the right side.

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The exact value of $\arcsin (\frac{1}{2})$ is $30$ .

$θ = 30$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$θ = 180 - 30$

Subtract $30$ from $180$ .

$θ = 150$

Find the period of $\sin (θ)$ .

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

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

$\frac{360}{| 1 |}$

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

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 30 + 360 n, 150 + 360 n$ , for any integer $n$

Set $\sin (θ) - 1$ equal to $0$ and solve for $θ$ .

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Set $\sin (θ) - 1$ equal to $0$ .

$\sin (θ) - 1 = 0$

Solve $\sin (θ) - 1 = 0$ for $θ$ .

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Add $1$ to both sides of the equation.

$\sin (θ) = 1$

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

$θ = \arcsin (1)$

Simplify the right side.

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The exact value of $\arcsin (1)$ is $90$ .

$θ = 90$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$θ = 180 - 90$

Subtract $90$ from $180$ .

$θ = 90$

Find the period of $\sin (θ)$ .

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

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

$\frac{360}{| 1 |}$

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

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 90 + 360 n$ , for any integer $n$

The final solution is all the values that make $(2 \sin (θ) - 1) (\sin (θ) - 1) = 0$ true.

$θ = 30 + 360 n, 150 + 360 n, 90 + 360 n$ , for any integer $n$

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