Solve for θ sin(theta)cos(theta)-1/2cos(theta)=0

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

Step 1

Divide each term in the equation by $\cos (θ)$ .

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

Step 2

Cancel the common factor of $\cos (θ)$ .

Cancel the common factor.

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

Divide $\sin (θ)$ by $1$ .

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

Step 3

Cancel the common factor of $\cos (θ)$ .

Cancel the common factor.

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

Divide $- \frac{1}{2}$ by $1$ .

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

Step 4

Separate fractions.

$\sin (θ) - \frac{1}{2} = \frac{0}{1} \cdot \frac{1}{\cos (θ)}$

Step 5

Convert from $\frac{1}{\cos (θ)}$ to $\sec (θ)$ .

$\sin (θ) - \frac{1}{2} = \frac{0}{1} \cdot \sec (θ)$

Step 6

Divide $0$ by $1$ .

$\sin (θ) - \frac{1}{2} = 0 \sec (θ)$

Step 7

Multiply $0$ by $\sec (θ)$ .

$\sin (θ) - \frac{1}{2} = 0$

Step 8

Add $\frac{1}{2}$ to both sides of the equation.

$\sin (θ) = \frac{1}{2}$

Step 9

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

$θ = \arcsin (\frac{1}{2})$

Step 10

Simplify the right side.

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

$θ = \frac{π}{6}$

Step 11

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.

$θ = π - \frac{π}{6}$

Step 12

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 $6 π$ .

$θ = \frac{5 π}{6}$

Step 13

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 $1$ .

$2 π$

Step 14

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 sin(theta)cos(theta)-1/2cos(theta)=0

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