Algebra Examples

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

Step 1

Subtract $\sin (x)$ from both sides of the equation.

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

Step 2

Replace $\cos^{2} (x)$ with $1 - \sin^{2} (x)$ .

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

Step 3

Solve for $x$ .

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Step 3.1

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

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

Step 3.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.3

Substitute the values $a = - 1$ , $b = - 1$ , and $c = 1$ into the quadratic formula and solve for $u$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (- 1 \cdot 1)}}{2 \cdot - 1}$

Step 3.4

Simplify.

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Step 3.4.1

Simplify the numerator.

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Step 3.4.1.1

Raise $- 1$ to the power of $2$ .

$u = \frac{1 \pm \sqrt{1 - 4 \cdot - 1 \cdot 1}}{2 \cdot - 1}$

Step 3.4.1.2

Multiply $- 4 \cdot - 1 \cdot 1$ .

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Step 3.4.1.2.1

Multiply $- 4$ by $- 1$ .

$u = \frac{1 \pm \sqrt{1 + 4 \cdot 1}}{2 \cdot - 1}$

Step 3.4.1.2.2

Multiply $4$ by $1$ .

$u = \frac{1 \pm \sqrt{1 + 4}}{2 \cdot - 1}$

Step 3.4.1.3

Add $1$ and $4$ .

$u = \frac{1 \pm \sqrt{5}}{2 \cdot - 1}$

Step 3.4.2

Multiply $2$ by $- 1$ .

$u = \frac{1 \pm \sqrt{5}}{- 2}$

Step 3.4.3

Move the negative in front of the fraction.

$u = - \frac{1 \pm \sqrt{5}}{2}$

Step 3.5

The final answer is the combination of both solutions.

$u = - \frac{1 + \sqrt{5}}{2}, - \frac{1 - \sqrt{5}}{2}$

Step 3.6

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

$\sin (x) = - \frac{1 + \sqrt{5}}{2}, - \frac{1 - \sqrt{5}}{2}$

Step 3.7

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

$\sin (x) = - \frac{1 + \sqrt{5}}{2}$

$\sin (x) = - \frac{1 - \sqrt{5}}{2}$

Step 3.8

Solve for $x$ in $\sin (x) = - \frac{1 + \sqrt{5}}{2}$ .

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Step 3.8.1

The range of sine is $- 1 \leq y \leq 1$ . Since $- \frac{1 + \sqrt{5}}{2}$ does not fall in this range, there is no solution.

No solution

Step 3.9

Solve for $x$ in $\sin (x) = - \frac{1 - \sqrt{5}}{2}$ .

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Step 3.9.1

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

$x = \arcsin (- \frac{1 - \sqrt{5}}{2})$

Step 3.9.2

Simplify the right side.

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Step 3.9.2.1

Evaluate $\arcsin (- \frac{1 - \sqrt{5}}{2})$ .

$x = 0.66623943$

Step 3.9.3

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 (3.14159265) - 0.66623943 + 3.14159265$

Step 3.9.4

Simplify the expression to find the second solution.

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Step 3.9.4.1

Subtract $2 π$ from $2 (3.14159265) - 0.66623943 + 3.14159265$ .

$x = 2 (3.14159265) - 0.66623943 + 3.14159265 - 2 π$

Step 3.9.4.2

The resulting angle of $2.47535322$ is positive, less than $2 π$ , and coterminal with $2 (3.14159265) - 0.66623943 + 3.14159265$ .

$x = 2.47535322$

Step 3.9.5

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

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Step 3.9.5.1

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

$\frac{2 π}{| b |}$

Step 3.9.5.2

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

$\frac{2 π}{| 1 |}$

Step 3.9.5.3

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

$\frac{2 π}{1}$

Step 3.9.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.9.6

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

$x = 0.66623943 + 2 π n, 2.47535322 + 2 π n$ , for any integer $n$

Step 3.10

List all of the solutions.

$x = 0.66623943 + 2 π n, 2.47535322 + 2 π n$ , for any integer $n$