Solve for θ 4sin(theta)^2-4sin(theta)+1=0

$4 \sin^{2} (θ) - 4 \sin (θ) + 1 = 0$

Substitute $u$ for $\sin (θ)$ .

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

Factor using the perfect square rule.

Rewrite $4 u^{2}$ as ${(2 u)}^{2}$ .

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

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

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

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

$4 u = 2 \cdot (2 u) \cdot 1$

Rewrite the polynomial.

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

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

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

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

$2 u - 1 = 0$

Solve 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 $1$ .

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

Substitute $\sin (θ)$ for $u$ .

$\sin (θ) = \frac{1}{2}$

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.

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

$θ = \frac{π}{6}$

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.

$θ = π - \frac{π}{6}$

Simplify $π - \frac{π}{6}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$θ = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{6}$

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 $1$ .

$2 π$

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

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

Solve for θ 4sin(theta)^2-4sin(theta)+1=0

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