Find All Complex Solutions square root of 2sin(theta)^2-sin(theta)=0

$\sqrt{2} \sin^{2} (θ) - \sin (θ) = 0$

Step 1

Factor $\sin (θ)$ out of $\sqrt{2} \sin^{2} (θ) - \sin (θ)$ .

Factor $\sin (θ)$ out of $\sqrt{2} \sin^{2} (θ)$ .

$\sin (θ) (\sqrt{2} \sin (θ)) - \sin (θ) = 0$

Factor $\sin (θ)$ out of $- \sin (θ)$ .

$\sin (θ) (\sqrt{2} \sin (θ)) + \sin (θ) \cdot - 1 = 0$

Factor $\sin (θ)$ out of $\sin (θ) (\sqrt{2} \sin (θ)) + \sin (θ) \cdot - 1$ .

$\sin (θ) (\sqrt{2} \sin (θ) - 1) = 0$

Step 2

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

$\sin (θ) = 0$

$\sqrt{2} \sin (θ) - 1 = 0$

Step 3

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

Set $\sin (θ)$ equal to $0$ .

$\sin (θ) = 0$

Solve $\sin (θ) = 0$ for $θ$ .

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

$θ = \arcsin (0)$

Simplify the right side.

The exact value of $\arcsin (0)$ is $0$ .

$θ = 0$

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.

$θ = π - 0$

Subtract $0$ from $π$ .

$θ = π$

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.

$θ = 2 π n, π + 2 π n$ , for any integer $n$

Step 4

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

Set $\sqrt{2} \sin (θ) - 1$ equal to $0$ .

$\sqrt{2} \sin (θ) - 1 = 0$

Solve $\sqrt{2} \sin (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$\sqrt{2} \sin (θ) = 1$

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

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

$\frac{\sqrt{2} \sin (θ)}{\sqrt{2}} = \frac{1}{\sqrt{2}}$

Simplify the left side.

Cancel the common factor of $\sqrt{2}$ .

Cancel the common factor.

$\frac{\sqrt{2} \sin (θ)}{\sqrt{2}} = \frac{1}{\sqrt{2}}$

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{1}{\sqrt{2}}$

Simplify the right side.

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\sin (θ) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\sin (θ) = \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$\sin (θ) = \frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

$\sin (θ) = \frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\sin (θ) = \frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Add $1$ and $1$ .

$\sin (θ) = \frac{\sqrt{2}}{{\sqrt{2}}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$\sin (θ) = \frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sin (θ) = \frac{\sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\sin (θ) = \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\sin (θ) = \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

$\sin (θ) = \frac{\sqrt{2}}{2^{1}}$

Evaluate the exponent.

$\sin (θ) = \frac{\sqrt{2}}{2}$

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

$θ = \arcsin (\frac{\sqrt{2}}{2})$

Simplify the right side.

The exact value of $\arcsin (\frac{\sqrt{2}}{2})$ is $\frac{π}{4}$ .

$θ = \frac{π}{4}$

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.

$θ = π - \frac{π}{4}$

Simplify $π - \frac{π}{4}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$θ = π \cdot \frac{4}{4} - \frac{π}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$θ = \frac{π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 4 - π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$θ = \frac{4 \cdot π - π}{4}$

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{4}$

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.

$θ = \frac{π}{4} + 2 π n, \frac{3 π}{4} + 2 π n$ , for any integer $n$

Step 5

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

$θ = 2 π n, π + 2 π n, \frac{π}{4} + 2 π n, \frac{3 π}{4} + 2 π n$ , for any integer $n$

Find All Complex Solutions square root of 2sin(theta)^2-sin(theta)=0

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