Trigonometry Examples

$\sin (x) + 1 = \sqrt{3} \cos (x)$

Subtract $1$ from both sides of the equation.

$\sin (x) = \sqrt{3} \cos (x) - 1$

Square both sides of the equation.

$\sin^{2} (x) = {(\sqrt{3} \cos (x) - 1)}^{2}$

Simplify ${(\sqrt{3} \cos (x) - 1)}^{2}$ .

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Rewrite ${(\sqrt{3} \cos (x) - 1)}^{2}$ as $(\sqrt{3} \cos (x) - 1) (\sqrt{3} \cos (x) - 1)$ .

$\sin^{2} (x) = (\sqrt{3} \cos (x) - 1) (\sqrt{3} \cos (x) - 1)$

Expand $(\sqrt{3} \cos (x) - 1) (\sqrt{3} \cos (x) - 1)$ using the FOIL Method.

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Apply the distributive property.

$\sin^{2} (x) = \sqrt{3} \cos (x) (\sqrt{3} \cos (x) - 1) - 1 (\sqrt{3} \cos (x) - 1)$

Apply the distributive property.

$\sin^{2} (x) = \sqrt{3} \cos (x) (\sqrt{3} \cos (x)) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x) - 1)$

Apply the distributive property.

$\sin^{2} (x) = \sqrt{3} \cos (x) (\sqrt{3} \cos (x)) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Simplify and combine like terms.

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Simplify each term.

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Multiply $\sqrt{3} \cos (x) (\sqrt{3} \cos (x))$ .

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Raise $\sqrt{3}$ to the power of $1$ .

$\sin^{2} (x) = {\sqrt{3}}^{1} \sqrt{3} \cos (x) \cos (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Raise $\sqrt{3}$ to the power of $1$ .

$\sin^{2} (x) = {\sqrt{3}}^{1} {\sqrt{3}}^{1} \cos (x) \cos (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\sin^{2} (x) = {\sqrt{3}}^{1 + 1} \cos (x) \cos (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Add $1$ and $1$ .

$\sin^{2} (x) = {\sqrt{3}}^{2} \cos (x) \cos (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Raise $\cos (x)$ to the power of $1$ .

$\sin^{2} (x) = {\sqrt{3}}^{2} (\cos^{1} (x) \cos (x)) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Raise $\cos (x)$ to the power of $1$ .

$\sin^{2} (x) = {\sqrt{3}}^{2} (\cos^{1} (x) \cos^{1} (x)) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\sin^{2} (x) = {\sqrt{3}}^{2} {\cos (x)}^{1 + 1} + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Add $1$ and $1$ .

$\sin^{2} (x) = {\sqrt{3}}^{2} \cos^{2} (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$\sin^{2} (x) = {(3^{\frac{1}{2}})}^{2} \cos^{2} (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sin^{2} (x) = 3^{\frac{1}{2} \cdot 2} \cos^{2} (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Combine $\frac{1}{2}$ and $2$ .

$\sin^{2} (x) = 3^{\frac{2}{2}} \cos^{2} (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\sin^{2} (x) = 3^{\frac{2}{2}} \cos^{2} (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Rewrite the expression.

$\sin^{2} (x) = 3^{1} \cos^{2} (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Evaluate the exponent.

$\sin^{2} (x) = 3 \cos^{2} (x) + \sqrt{3} \cos (x) \cdot - 1 - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Move $- 1$ to the left of $\sqrt{3} \cos (x)$ .

$\sin^{2} (x) = 3 \cos^{2} (x) - 1 \cdot (\sqrt{3} \cos (x)) - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Rewrite $- 1 (\sqrt{3} \cos (x))$ as $- (\sqrt{3} \cos (x))$ .

$\sin^{2} (x) = 3 \cos^{2} (x) - (\sqrt{3} \cos (x)) - 1 (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Rewrite $- 1 (\sqrt{3} \cos (x))$ as $- (\sqrt{3} \cos (x))$ .

$\sin^{2} (x) = 3 \cos^{2} (x) - \sqrt{3} \cos (x) - (\sqrt{3} \cos (x)) - 1 \cdot - 1$

Multiply $- 1$ by $- 1$ .

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

Subtract $\sqrt{3} \cos (x)$ from $- \sqrt{3} \cos (x)$ .

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

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

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Subtract $3 \cos^{2} (x)$ from both sides of the equation.

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

Add $2 \sqrt{3} \cos (x)$ to both sides of the equation.

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

Subtract $1$ from both sides of the equation.

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

Simplify $\sin^{2} (x) - 3 \cos^{2} (x) + 2 \sqrt{3} \cos (x) - 1$ .

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Move $- 1$ .

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

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

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

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

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

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

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

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

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

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

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

Apply pythagorean identity.

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

Subtract $3 \cos^{2} (x)$ from $- \cos^{2} (x)$ .

$- 4 \cos^{2} (x) + 2 \sqrt{3} \cos (x) = 0$

Solve for $x$ .

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Factor $2 \cos (x)$ out of $- 4 \cos^{2} (x) + 2 \sqrt{3} \cos (x)$ .

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Factor $2 \cos (x)$ out of $- 4 \cos^{2} (x)$ .

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

Factor $2 \cos (x)$ out of $2 \sqrt{3} \cos (x)$ .

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

Factor $2 \cos (x)$ out of $2 \cos (x) (- 2 \cos (x)) + 2 \cos (x) \sqrt{3}$ .

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

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

$\cos (x) = 0$

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

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

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

$\cos (x) = 0$

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

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

$x = \arccos (0)$

Simplify the right side.

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The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

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.

$x = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Combine fractions.

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Combine $2 π$ and $\frac{2}{2}$ .

$x = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

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Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

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.

$x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Set $- 2 \cos (x) + \sqrt{3}$ equal to $0$ and solve for $x$ .

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Set $- 2 \cos (x) + \sqrt{3}$ equal to $0$ .

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

Solve $- 2 \cos (x) + \sqrt{3} = 0$ for $x$ .

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Subtract $\sqrt{3}$ from both sides of the equation.

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

Divide each term in $- 2 \cos (x) = - \sqrt{3}$ by $- 2$ and simplify.

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Divide each term in $- 2 \cos (x) = - \sqrt{3}$ by $- 2$ .

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

Simplify the left side.

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Cancel the common factor of $- 2$ .

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Cancel the common factor.

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

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

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

Simplify the right side.

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Dividing two negative values results in a positive value.

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

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

$x = \arccos (\frac{\sqrt{3}}{2})$

Simplify the right side.

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The exact value of $\arccos (\frac{\sqrt{3}}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

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.

$x = 2 π - \frac{π}{6}$

Simplify $2 π - \frac{π}{6}$ .

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To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = 2 π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

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Combine $2 π$ and $\frac{6}{6}$ .

$x = \frac{2 π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 6 - π}{6}$

Simplify the numerator.

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Multiply $6$ by $2$ .

$x = \frac{12 π - π}{6}$

Subtract $π$ from $12 π$ .

$x = \frac{11 π}{6}$

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.

$x = \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

The final solution is all the values that make $2 \cos (x) (- 2 \cos (x) + \sqrt{3}) = 0$ true.

$x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Consolidate $\frac{π}{2} + 2 π n$ and $\frac{3 π}{2} + 2 π n$ to $\frac{π}{2} + π n$ .

$x = \frac{π}{2} + π n, \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Exclude the solutions that do not make $\sin (x) + 1 = \sqrt{3} \cos (x)$ true.

$x = 0.52359877, 4.71238898, 6.80678408, 10.99557428, 13.08996938, 17.27875959, 19.37315469, 23.5619449, 25.65634, 31.93952531, 38.22271061, 44.50589592$ , for any integer $n$