Solve for x (tan(x)+ square root of 3)(2sin(x)- square root of 2)=0

$(\tan (x) + \sqrt{3}) (2 \sin (x) - \sqrt{2}) = 0$

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

$\tan (x) + \sqrt{3} = 0$

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

Set $\tan (x) + \sqrt{3}$ equal to $0$ and solve for $x$ .

Set $\tan (x) + \sqrt{3}$ equal to $0$ .

$\tan (x) + \sqrt{3} = 0$

Solve $\tan (x) + \sqrt{3} = 0$ for $x$ .

Subtract $\sqrt{3}$ from both sides of the equation.

$\tan (x) = - \sqrt{3}$

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

$x = \arctan (- \sqrt{3})$

Simplify the right side.

The exact value of $\arctan (- \sqrt{3})$ is $- \frac{π}{3}$ .

$x = - \frac{π}{3}$

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

$x = - \frac{π}{3} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $- \frac{π}{3} - π$ .

$x = - \frac{π}{3} - π + 2 π$

The resulting angle of $\frac{2 π}{3}$ is positive and coterminal with $- \frac{π}{3} - π$ .

$x = \frac{2 π}{3}$

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 $1$ .

$π$

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

Add $π$ to $- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + π$

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

$π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

Combine $π$ and $\frac{3}{3}$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$\frac{π \cdot 3 - π}{3}$

Simplify the numerator.

Move $3$ to the left of $π$ .

$\frac{3 \cdot π - π}{3}$

Subtract $π$ from $3 π$ .

$\frac{2 π}{3}$

List the new angles.

$x = \frac{2 π}{3}$

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

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

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

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

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

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

Add $\sqrt{2}$ to both sides of the equation.

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

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

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 $n$

The final solution is all the values that make $(\tan (x) + \sqrt{3}) (2 \sin (x) - \sqrt{2}) = 0$ true.

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

Solve for x (tan(x)+ square root of 3)(2sin(x)- square root of 2)=0

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