Solve for ? two sin(x) to the power of two minus three sin(x) plus one equals zero

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

The left part of the equation should be factored.

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

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

Apply the grouping factorization.

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For a polynomial of the form $a x^{2} + b x + c$ , Should be rewritten 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 $- 3 u$ .

$2 u^{2} - 3 (u) + 1$

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

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

Use the distributive law.

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

Take the greatest common factor from each group out of the brackets.

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Put together the Initially two terms and the last two terms.

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

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

$2 u (u - 1) - 1 (u - 1)$

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

$(u - 1) (2 u - 1)$

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

$(\sin (x) - 1) (2 \sin (x) - 1)$

Replace the left side with the factored expression.

$(\sin (x) - 1) (2 \sin (x) - 1) = 0$

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 $- 3 \sin (x)$ .

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

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

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

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

$2 \sin (x) (\sin (x) - 1) - 1 (\sin (x) - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $\sin (x) - 1$ .

$(\sin (x) - 1) (2 \sin (x) - 1) = 0$

In case any individual factor in the left part of the equation equals $0$ , the entire expression will be equal to $0$ .

$\sin (x) - 1 = 0$

$2 \sin (x) - 1 = 0$

Make the initial factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$\sin (x) - 1 = 0$

Add $1$ to both sides of the equation.

$\sin (x) = 1$

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

$x = \arcsin (1)$

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

$x = \frac{π}{2}$

The 1st and 2nd quadrants of the sine function are positive. To calculate the second answer, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - \frac{π}{2}$

Clarify $π - \frac{π}{2}$ .

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To write $\frac{π}{1}$ as a fraction with a common denominator, First, multiply by $\frac{2}{2}$ .

$x = \frac{π}{1} \cdot \frac{2}{2} - \frac{π}{2}$

Write each expression with a common denominator of $2$ , by multiplying each by an appropriate factor of $1$ .

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Add.

$x = \frac{π \cdot 2}{1 \cdot 2} - \frac{π}{2}$

First, multiply $2$ by $1$ .

$x = \frac{π \cdot 2}{2} - \frac{π}{2}$

Add numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Clarify the numerator.

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Put $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Calculate the period.

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Evaluate the period of function using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

Find solution to the equation.

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

$\frac{2 π}{1}$

First, 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 two directions at once.

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

Make the following factor equal to $0$ and solve.

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Set the Then factor equal to $0$ .

$2 \sin (x) - 1 = 0$

Add $1$ to both sides of the equation.

$2 \sin (x) = 1$

Divide each term by $2$ and simplify.

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

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

The common factor should be canceled of $2$ .

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Cancel the common factor.

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

Divide $\sin (x)$ by $1$ .

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

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

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

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 calculate the second answer, subtract the reference angle from $π$ to find the solution in the second quadrant.

$x = π - \frac{π}{6}$

Simplify $π - \frac{π}{6}$ .

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To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = \frac{π}{1} \cdot \frac{6}{6} - \frac{π}{6}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{π \cdot 6}{1 \cdot 6} - \frac{π}{6}$

Multiply $6$ by $1$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

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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.

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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.

$\frac{2 π}{| 1 |}$

Solve the equation.

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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 $n$

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

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

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