Evaluate sin(theta)+cos(theta)*cot(theta)=2

$\sin (θ) + \cos (θ) \cdot \cot (θ) = 2$

Simplify the left side.

Simplify each term.

Rewrite $\cot (θ)$ in terms of sines and cosines.

$\sin (θ) + \frac{\cos (θ) \cos (θ)}{\sin (θ)} = 2$

Multiply $\cos (θ) \frac{\cos (θ)}{\sin (θ)}$ .

Raise $\cos (θ)$ to the power of $1$ .

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

Raise $\cos (θ)$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

Multiply both sides of the equation by $\sin (θ)$ .

$\sin (θ) (\sin (θ) + \frac{\cos^{2} (θ)}{\sin (θ)}) = \sin (θ) \cdot 2$

Apply the distributive property.

$\sin (θ) \sin (θ) + \sin (θ) \frac{\cos^{2} (θ)}{\sin (θ)} = \sin (θ) \cdot 2$

Multiply $\sin (θ) \sin (θ)$ .

Raise $\sin (θ)$ to the power of $1$ .

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

Raise $\sin (θ)$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

$\sin^{2} (θ) + \sin (θ) \frac{\cos^{2} (θ)}{\sin (θ)} = \sin (θ) \cdot 2$

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

Cancel the common factor.

$\sin^{2} (θ) + \sin (θ) \frac{\cos^{2} (θ)}{\sin (θ)} = \sin (θ) \cdot 2$

Rewrite the expression.

$\sin^{2} (θ) + \cos^{2} (θ) = \sin (θ) \cdot 2$

Apply pythagorean identity.

$1 = \sin (θ) \cdot 2$

Move $2$ to the left of $\sin (θ)$ .

$1 = 2 \sin (θ)$

Rewrite the equation as $2 \sin (θ) = 1$ .

$2 \sin (θ) = 1$

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

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

$\frac{2 \sin (θ)}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \sin (θ)}{2} = \frac{1}{2}$

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{1}{2}$

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

$θ = \arcsin (\frac{1}{2})$

Simplify the right side.

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

$θ = \frac{π}{6}$

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.

$θ = π - \frac{π}{6}$

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 $6 π$ .

$θ = \frac{5 π}{6}$

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.

$θ = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Evaluate sin(theta)+cos(theta)*cot(theta)=2

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