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

$\sin^{2} (θ) - 2 \sin (θ) + 1 = 0$

Step 1

Substitute $u$ for $\sin (θ)$ .

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

Step 2

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 1$ .

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

Step 3

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

$u - 1 = 0$

Step 4

Add $1$ to both sides of the equation.

$u = 1$

Step 5

Substitute $\sin (θ)$ for $u$ .

$\sin (θ) = 1$

Step 6

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

$θ = \arcsin (1)$

Step 7

Simplify the right side.

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

$θ = \frac{π}{2}$

Step 8

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.

$θ = π - \frac{π}{2}$

Step 9

Simplify $π - \frac{π}{2}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 2 - π}{2}$

Simplify the numerator.

Move $2$ to the left of $π$ .

$θ = \frac{2 \cdot π - π}{2}$

Subtract $π$ from $2 π$ .

$θ = \frac{π}{2}$

Step 10

Find the period of $\sin (θ)$ .

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 $1$ .

$2 π$

Step 11

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

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

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

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