Trigonometry Examples

$3 \cos^{2} (x) + 2 \cos (x) - 1 = 0$

Factor the left side of the equation.

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

$3 u^{2} + 2 u - 1$

Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 1 = - 3$ and whose sum is $b = 2$ .

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Factor $2$ out of $2 u$ .

$3 u^{2} + 2 (u) - 1$

Rewrite $2$ as $- 1$ plus $3$

$3 u^{2} + (- 1 + 3) u - 1$

Apply the distributive property.

$3 u^{2} - 1 u + 3 u - 1$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(3 u^{2} - 1 u) + 3 u - 1$

Factor out the greatest common factor (GCF) from each group.

$u (3 u - 1) + 1 (3 u - 1)$

Factor the polynomial by factoring out the greatest common factor, $3 u - 1$ .

$(3 u - 1) (u + 1)$

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

$(3 \cos (x) - 1) (\cos (x) + 1)$

Replace the left side with the factored expression.

$(3 \cos (x) - 1) (\cos (x) + 1) = 0$

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

$3 \cos (x) - 1 = 0$

$\cos (x) + 1 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$3 \cos (x) - 1 = 0$

Add $1$ to both sides of the equation.

$3 \cos (x) = 1$

Divide each term by $3$ and simplify.

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Divide each term in $3 \cos (x) = 1$ by $3$ .

$\frac{3 \cos (x)}{3} = \frac{1}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 \cos (x)}{3} = \frac{1}{3}$

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

$\cos (x) = \frac{1}{3}$

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

$x = \arccos (\frac{1}{3})$

Evaluate $\arccos (\frac{1}{3})$ .

$x = 1.23095941$

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

$x = 2 (3.14159265) - 1.23095941$

Simplify the expression to find the second solution.

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Remove parentheses.

$x = 2 (3.14159265) - 1.23095941$

Simplify $2 (3.14159265) - 1.23095941$ .

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Multiply $2$ by $3.14159265$ .

$x = 6.2831853 - 1.23095941$

Subtract $1.23095941$ from $6.2831853$ .

$x = 5.05222588$

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

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

$x = 1.23095941 + 2 π n, 5.05222588 + 2 π n$ , for any integer $n$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$\cos (x) + 1 = 0$

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

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

$x = \arccos (- 1)$

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

$x = π$

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.

$x = 2 π - π$

Subtract $π$ from $2 π$ .

$x = π$

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

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

$x = π + 2 π n$ , for any integer $n$

The final solution is all the values that make $(3 \cos (x) - 1) (\cos (x) + 1) = 0$ true.

$x = 1.23095941 + 2 π n, 5.05222588 + 2 π n, π + 2 π n$ , for any integer $n$

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