Trigonometry Examples

$- 4 \sin (x) = - \cos^{2} (x) + 1$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Add $\cos^{2} (x)$ to both sides of the equation.

$- 4 \sin (x) + \cos^{2} (x) = 1$

Step 1.2

Subtract $1$ from both sides of the equation.

$- 4 \sin (x) + \cos^{2} (x) - 1 = 0$

Step 2

Simplify $- 4 \sin (x) + \cos^{2} (x) - 1$ .

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Step 2.1

Reorder $\cos^{2} (x)$ and $- 1$ .

$- 4 \sin (x) - 1 + \cos^{2} (x) = 0$

Step 2.2

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

$- 4 \sin (x) - 1 (1) + \cos^{2} (x) = 0$

Step 2.3

Factor $- 1$ out of $\cos^{2} (x)$ .

$- 4 \sin (x) - 1 (1) - 1 (- \cos^{2} (x)) = 0$

Step 2.4

Factor $- 1$ out of $- 1 (1) - 1 (- \cos^{2} (x))$ .

$- 4 \sin (x) - 1 (1 - \cos^{2} (x)) = 0$

Step 2.5

Rewrite $- 1 (1 - \cos^{2} (x))$ as $- (1 - \cos^{2} (x))$ .

$- 4 \sin (x) - (1 - \cos^{2} (x)) = 0$

Step 2.6

Apply pythagorean identity.

$- 4 \sin (x) - \sin^{2} (x) = 0$

Step 3

Solve for $x$ .

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Step 3.1

Factor the left side of the equation.

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Step 3.1.1

Let $u = \sin (x)$ . Substitute $u$ for all occurrences of $\sin (x)$ .

$- 4 u - u^{2} = 0$

Step 3.1.2

Factor $u$ out of $- 4 u - u^{2}$ .

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Step 3.1.2.1

Factor $u$ out of $- 4 u$ .

$u \cdot - 4 - u^{2} = 0$

Step 3.1.2.2

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

$u \cdot - 4 + u (- u) = 0$

Step 3.1.2.3

Factor $u$ out of $u \cdot - 4 + u (- u)$ .

$u (- 4 - u) = 0$

Step 3.1.3

Replace all occurrences of $u$ with $\sin (x)$ .

$\sin (x) (- 4 - \sin (x)) = 0$

Step 3.2

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

$\sin (x) = 0$

$- 4 - \sin (x) = 0$

Step 3.3

Set $\sin (x)$ equal to $0$ and solve for $x$ .

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Step 3.3.1

Set $\sin (x)$ equal to $0$ .

$\sin (x) = 0$

Step 3.3.2

Solve $\sin (x) = 0$ for $x$ .

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Step 3.3.2.1

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

$x = \arcsin (0)$

Step 3.3.2.2

Simplify the right side.

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Step 3.3.2.2.1

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

Step 3.3.2.3

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.

$x = π - 0$

Step 3.3.2.4

Subtract $0$ from $π$ .

$x = π$

Step 3.3.2.5

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

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Step 3.3.2.5.1

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

$\frac{2 π}{| b |}$

Step 3.3.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 3.3.2.5.3

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

$\frac{2 π}{1}$

Step 3.3.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.3.2.6

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

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 3.4

Set $- 4 - \sin (x)$ equal to $0$ and solve for $x$ .

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Step 3.4.1

Set $- 4 - \sin (x)$ equal to $0$ .

$- 4 - \sin (x) = 0$

Step 3.4.2

Solve $- 4 - \sin (x) = 0$ for $x$ .

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Step 3.4.2.1

Add $4$ to both sides of the equation.

$- \sin (x) = 4$

Step 3.4.2.2

Divide each term in $- \sin (x) = 4$ by $- 1$ and simplify.

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Step 3.4.2.2.1

Divide each term in $- \sin (x) = 4$ by $- 1$ .

$\frac{- \sin (x)}{- 1} = \frac{4}{- 1}$

Step 3.4.2.2.2

Simplify the left side.

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Step 3.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (x)}{1} = \frac{4}{- 1}$

Step 3.4.2.2.2.2

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

$\sin (x) = \frac{4}{- 1}$

Step 3.4.2.2.3

Simplify the right side.

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Step 3.4.2.2.3.1

Divide $4$ by $- 1$ .

$\sin (x) = - 4$

Step 3.4.2.3

The range of sine is $- 1 \leq y \leq 1$ . Since $- 4$ does not fall in this range, there is no solution.

No solution

Step 3.5

The final solution is all the values that make $\sin (x) (- 4 - \sin (x)) = 0$ true.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 4

Consolidate the answers.

$x = π n$ , for any integer $n$