Solve for θ 2sin(theta)+4=4

$2 \sin (θ) + 4 = 4$

Move all terms not containing $\sin (θ)$ to the right side of the equation.

Subtract $4$ from both sides of the equation.

$2 \sin (θ) = 4 - 4$

Subtract $4$ from $4$ .

$2 \sin (θ) = 0$

Divide each term in $2 \sin (θ) = 0$ by $2$ and simplify.

Divide each term in $2 \sin (θ) = 0$ by $2$ .

$\frac{2 \sin (θ)}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \sin (θ)}{2} = \frac{0}{2}$

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$\sin (θ) = 0$

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 $n$

Consolidate the answers.

$θ = π n$ , for any integer $n$

Solve for θ 2sin(theta)+4=4

Related Questions

Support LCMGCF.com by disabling your adblocker.