Precalculus Examples

$\cos (2 x) = \frac{\sqrt{3}}{2}$ , $(0, 2 π)$

Step 1

Subtract $\cos (2 x)$ from both sides of the equation.

$0 = \frac{\sqrt{3}}{2} - \cos (2 x)$

Step 2

The Intermediate Value Theorem states that, if $f$ is a real-valued continuous function on the interval $[a, b]$ , and $u$ is a number between $f (a)$ and $f (b)$ , then there is a $c$ contained in the interval $[a, b]$ such that $f (c) = u$ .

$u = f (c) = 0$

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

Simplify each term.

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Step 4.1

Multiply $2$ by $0$ .

$f (0) = \frac{\sqrt{3}}{2} - \cos (0)$

Step 4.2

The exact value of $\cos (0)$ is $1$ .

$f (0) = \frac{\sqrt{3}}{2} - 1 \cdot 1$

Step 4.3

Multiply $- 1$ by $1$ .

$f (0) = \frac{\sqrt{3}}{2} - 1$

Step 5

Simplify each term.

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Step 5.1

Multiply $2$ by $2$ .

$f (2 π) = \frac{\sqrt{3}}{2} - \cos (4 π)$

Step 5.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$f (2 π) = \frac{\sqrt{3}}{2} - \cos (0)$

Step 5.3

The exact value of $\cos (0)$ is $1$ .

$f (2 π) = \frac{\sqrt{3}}{2} - 1 \cdot 1$

Step 5.4

Multiply $- 1$ by $1$ .

$f (2 π) = \frac{\sqrt{3}}{2} - 1$

Step 6

$0$ is not on the interval $[\frac{\sqrt{3}}{2} - 1, \frac{\sqrt{3}}{2} - 1]$ .

There is no root on the interval.

Step 7