Trigonometry Examples

$\sin^{2} (x) = 6 (\cos (x) + 1)$

Step 1

Replace the $\sin^{2} (x)$ with $1 - \cos^{2} (x)$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$1 - \cos^{2} (x) = 6 (\cos (x) + 1)$

Step 2

Reorder the polynomial.

$- \cos^{2} (x) + 1 = 6 (\cos (x) + 1)$

Step 3

Substitute $u$ for $\cos (x)$ .

$- {(u)}^{2} + 1 = 6 ((u) + 1)$

Step 4

Simplify $6 (u + 1)$ .

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Step 4.1

Rewrite.

$- u^{2} + 1 = 0 + 0 + 6 (u + 1)$

Step 4.2

Simplify by adding zeros.

$- u^{2} + 1 = 6 (u + 1)$

Step 4.3

Apply the distributive property.

$- u^{2} + 1 = 6 u + 6 \cdot 1$

Step 4.4

Multiply $6$ by $1$ .

$- u^{2} + 1 = 6 u + 6$

Step 5

Subtract $6 u$ from both sides of the equation.

$- u^{2} + 1 - 6 u = 6$

Step 6

Subtract $6$ from both sides of the equation.

$- u^{2} + 1 - 6 u - 6 = 0$

Step 7

Subtract $6$ from $1$ .

$- u^{2} - 6 u - 5 = 0$

Step 8

Factor the left side of the equation.

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Step 8.1

Factor $- 1$ out of $- u^{2} - 6 u - 5$ .

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Step 8.1.1

Factor $- 1$ out of $- u^{2}$ .

$- u^{2} - 6 u - 5 = 0$

Step 8.1.2

Factor $- 1$ out of $- 6 u$ .

$- u^{2} - (6 u) - 5 = 0$

Step 8.1.3

Rewrite $- 5$ as $- 1 (5)$ .

$- u^{2} - (6 u) - 1 \cdot 5 = 0$

Step 8.1.4

Factor $- 1$ out of $- u^{2} - (6 u)$ .

$- (u^{2} + 6 u) - 1 \cdot 5 = 0$

Step 8.1.5

Factor $- 1$ out of $- (u^{2} + 6 u) - 1 (5)$ .

$- (u^{2} + 6 u + 5) = 0$

Step 8.2

Factor.

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Step 8.2.1

Factor $u^{2} + 6 u + 5$ using the AC method.

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Step 8.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $5$ and whose sum is $6$ .

$1, 5$

Step 8.2.1.2

Write the factored form using these integers.

$- ((u + 1) (u + 5)) = 0$

Step 8.2.2

Remove unnecessary parentheses.

$- (u + 1) (u + 5) = 0$

Step 9

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

$u + 1 = 0$

$u + 5 = 0$

Step 10

Set $u + 1$ equal to $0$ and solve for $u$ .

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Step 10.1

Set $u + 1$ equal to $0$ .

$u + 1 = 0$

Step 10.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 11

Set $u + 5$ equal to $0$ and solve for $u$ .

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Step 11.1

Set $u + 5$ equal to $0$ .

$u + 5 = 0$

Step 11.2

Subtract $5$ from both sides of the equation.

$u = - 5$

Step 12

The final solution is all the values that make $- (u + 1) (u + 5) = 0$ true.

$u = - 1, - 5$

Step 13

Substitute $\cos (x)$ for $u$ .

$\cos (x) = - 1, - 5$

Step 14

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

$\cos (x) = - 1$

$\cos (x) = - 5$

Step 15

Solve for $x$ in $\cos (x) = - 1$ .

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Step 15.1

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

$x = \arccos (- 1)$

Step 15.2

Simplify the right side.

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Step 15.2.1

The exact value of $\arccos (- 1)$ is $π$ .

$x = π$

Step 15.3

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - π$

Step 15.4

Subtract $π$ from $2 π$ .

$x = π$

Step 15.5

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

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Step 15.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 15.5.2

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

$\frac{2 π}{| 1 |}$

Step 15.5.3

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

$\frac{2 π}{1}$

Step 15.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 15.6

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

$x = π + 2 π n$ , for any integer $n$

Step 16

Solve for $x$ in $\cos (x) = - 5$ .

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Step 16.1

The range of cosine is $- 1 \leq y \leq 1$ . Since $- 5$ does not fall in this range, there is no solution.

No solution

Step 17

List all of the solutions.

$x = π + 2 π n$ , for any integer $n$