Solve Over the Interval 2sin(x)^2tan(x)=tan(x) , [0,2pi)

$2 \sin^{2} (x) \tan (x) = \tan (x)$ , $[0, 2 π)$

Step 1

Subtract $\tan (x)$ from both sides of the equation.

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

Step 2

Factor $\tan (x)$ out of $2 \sin^{2} (x) \tan (x) - \tan (x)$ .

Factor $\tan (x)$ out of $2 \sin^{2} (x) \tan (x)$ .

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

Factor $\tan (x)$ out of $- \tan (x)$ .

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

Factor $\tan (x)$ out of $\tan (x) (2 \sin^{2} (x)) + \tan (x) \cdot - 1$ .

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

Step 3

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

$\tan (x) = 0$

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

Step 4

Set $\tan (x)$ equal to $0$ and solve for $x$ .

Set $\tan (x)$ equal to $0$ .

$\tan (x) = 0$

Solve $\tan (x) = 0$ for $x$ .

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

$x = \arctan (0)$

Simplify the right side.

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

$x = 0$

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + 0$

Add $π$ and $0$ .

$x = π$

Find the period of $\tan (x)$ .

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

$x = π n, π + π n$ , for any integer $n$

Step 5

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

Set $2 \sin^{2} (x) - 1$ equal to $0$ .

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

Solve $2 \sin^{2} (x) - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

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

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $\sin^{2} (x)$ by $1$ .

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

Take the square root of both sides of the equation to eliminate the exponent on the left side.

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

Simplify $\pm \sqrt{\frac{1}{2}}$ .

Rewrite $\sqrt{\frac{1}{2}}$ as $\frac{\sqrt{1}}{\sqrt{2}}$ .

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

Any root of $1$ is $1$ .

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

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

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

Combine and simplify the denominator.

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

$\sin (x) = \pm \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

$\sin (x) = \pm \frac{\sqrt{2}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Next, use the negative value of the $\pm$ to find the second solution.

$\sin (x) = - \frac{\sqrt{2}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$\sin (x) = \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}$

Set up each of the solutions to solve for $x$ .

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

$\sin (x) = - \frac{\sqrt{2}}{2}$

Solve for $x$ in $\sin (x) = \frac{\sqrt{2}}{2}$ .

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

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

Simplify the right side.

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

$x = \frac{π}{4}$

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.

$x = π - \frac{π}{4}$

Simplify $π - \frac{π}{4}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 - π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π - π}{4}$

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{4}$

Find the period of $\sin (x)$ .

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 $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

Solve for $x$ in $\sin (x) = - \frac{\sqrt{2}}{2}$ .

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

$x = \arcsin (- \frac{\sqrt{2}}{2})$

Simplify the right side.

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

$x = - \frac{π}{4}$

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{4} + π$

Simplify the expression to find the second solution.

Subtract $2 π$ from $2 π + \frac{π}{4} + π$ .

$x = 2 π + \frac{π}{4} + π - 2 π$

The resulting angle of $\frac{5 π}{4}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{4} + π$ .

$x = \frac{5 π}{4}$

Find the period of $\sin (x)$ .

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 $1$ .

$2 π$

Add $2 π$ to every negative angle to get positive angles.

Add $2 π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + 2 π$

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$2 π \cdot \frac{4}{4} - \frac{π}{4}$

Combine fractions.

Combine $2 π$ and $\frac{4}{4}$ .

$\frac{2 π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 4 - π}{4}$

Simplify the numerator.

Multiply $4$ by $2$ .

$\frac{8 π - π}{4}$

Subtract $π$ from $8 π$ .

$\frac{7 π}{4}$

List the new angles.

$x = \frac{7 π}{4}$

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

$x = \frac{5 π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

List all of the solutions.

$x = \frac{π}{4} + 2 π n, \frac{3 π}{4} + 2 π n, \frac{5 π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

Consolidate the answers.

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

Step 6

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

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

Step 7

Consolidate $π n$ and $π + π n$ to $π n$ .

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

Step 8

Find the values of $n$ that produce a value within the interval $[0, 2 π)$ .

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $0$ for $n$ .

$π (0)$

Multiply $π$ by $0$ .

$0$

The interval $[0, 2 π)$ contains $0$ .

$x = 0$

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $0$ for $n$ .

$\frac{π}{4} + \frac{π (0)}{2}$

Simplify.

Simplify each term.

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $π (0)$ .

$\frac{π}{4} + \frac{2 (π \cdot (0))}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{π}{4} + \frac{2 (π \cdot (0))}{2 (1)}$

Cancel the common factor.

$\frac{π}{4} + \frac{2 (π \cdot (0))}{2 \cdot 1}$

Rewrite the expression.

$\frac{π}{4} + \frac{π \cdot (0)}{1}$

Divide $π \cdot (0)$ by $1$ .

$\frac{π}{4} + π \cdot (0)$

Multiply $π$ by $0$ .

$\frac{π}{4} + 0$

Add $\frac{π}{4}$ and $0$ .

$\frac{π}{4}$

The interval $[0, 2 π)$ contains $\frac{π}{4}$ .

$x = 0, \frac{π}{4}$

Plug in $1$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $1$ for $n$ .

$\frac{π}{4} + \frac{π (1)}{2}$

Simplify.

Multiply $π$ by $1$ .

$\frac{π}{4} + \frac{π}{2}$

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

$\frac{π}{4} + \frac{π}{2} \cdot \frac{2}{2}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{π}{2}$ by $\frac{2}{2}$ .

$\frac{π}{4} + \frac{π \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{π}{4} + \frac{π \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{π + π \cdot 2}{4}$

Simplify the numerator.

Move $2$ to the left of $π$ .

$\frac{π + 2 \cdot π}{4}$

Add $π$ and $2 π$ .

$\frac{3 π}{4}$

The interval $[0, 2 π)$ contains $\frac{3 π}{4}$ .

$x = 0, \frac{π}{4}, \frac{3 π}{4}$

Plug in $1$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $1$ for $n$ .

$π (1)$

Multiply $π$ by $1$ .

$π$

The interval $[0, 2 π)$ contains $π$ .

$x = 0, \frac{π}{4}, \frac{3 π}{4}, π$

Plug in $2$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $2$ for $n$ .

$\frac{π}{4} + \frac{π (2)}{2}$

Simplify.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{π}{4} + \frac{π \cdot 2}{2}$

Divide $π$ by $1$ .

$\frac{π}{4} + π$

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

$\frac{π}{4} + π \cdot \frac{4}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$\frac{π}{4} + \frac{π \cdot 4}{4}$

Combine the numerators over the common denominator.

$\frac{π + π \cdot 4}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$\frac{π + 4 \cdot π}{4}$

Add $π$ and $4 π$ .

$\frac{5 π}{4}$

The interval $[0, 2 π)$ contains $\frac{5 π}{4}$ .

$x = 0, \frac{π}{4}, \frac{3 π}{4}, π, \frac{5 π}{4}$

Plug in $3$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $3$ for $n$ .

$\frac{π}{4} + \frac{π (3)}{2}$

Simplify.

Move $3$ to the left of $π$ .

$\frac{π}{4} + \frac{3 π}{2}$

To write $\frac{3 π}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{π}{4} + \frac{3 π}{2} \cdot \frac{2}{2}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3 π}{2}$ by $\frac{2}{2}$ .

$\frac{π}{4} + \frac{3 π \cdot 2}{2 \cdot 2}$

Multiply $2$ by $2$ .

$\frac{π}{4} + \frac{3 π \cdot 2}{4}$

Combine the numerators over the common denominator.

$\frac{π + 3 π \cdot 2}{4}$

Simplify the numerator.

Multiply $2$ by $3$ .

$\frac{π + 6 π}{4}$

Add $π$ and $6 π$ .

$\frac{7 π}{4}$

The interval $[0, 2 π)$ contains $\frac{7 π}{4}$ .

$x = 0, \frac{π}{4}, \frac{3 π}{4}, π, \frac{5 π}{4}, \frac{7 π}{4}$

Solve Over the Interval 2sin(x)^2tan(x)=tan(x) , [0,2pi)

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