7.5 Solving trigonometric equations (Page 2/7)

Precalculus

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Solving a linear equation involving the sine function

Find all possible exact solutions for the equation $\sin t = \frac{1}{2} .$

Solving for all possible values of t means that solutions include angles beyond the period of $2 π .$ From [link] , we can see that the solutions are $\frac{π}{6}$ and $\frac{5 π}{6} .$ But the problem is asking for all possible values that solve the equation. Therefore, the answer is

\frac{π}{6} \pm 2 π k and \frac{5 π}{6} \pm 2 π k

where $k$ is an integer.

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How To

Given a trigonometric equation, solve using algebra .

Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
Substitute the trigonometric expression with a single variable, such as $x$ or $u .$
Solve the equation the same way an algebraic equation would be solved.
Substitute the trigonometric expression back in for the variable in the resulting expressions.
Solve for the angle.

Solve the trigonometric equation in linear form

Solve the equation exactly: $2 \cos θ - 3 = - 5, 0 \leq θ < 2 π .$

Use algebraic techniques to solve the equation.

\begin{array}{l} 2 \cos θ - 3 = - 5 \\ 2 \cos θ = - 2 \\ \cos θ = - 1 \\ θ = π \end{array}

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Try It

Solve exactly the following linear equation on the interval $[0, 2 π) : 2 \sin x + 1 = 0.$

$x = \frac{7 π}{6}, \frac{11 π}{6}$

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Solving equations involving a single trigonometric function

When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and the unit circle (see [link] ). We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the reciprocal function, and solve for the angles using the function. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is $π,$ not $2 π .$ Further, the domain of tangent is all real numbers with the exception of odd integer multiples of $\frac{π}{2},$ unless, of course, a problem places its own restrictions on the domain.

Solving a problem involving a single trigonometric function

Solve the problem exactly: $2 \sin^{2} θ - 1 = 0, 0 \leq θ < 2 π .$

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate $\sin θ .$ Then we will find the angles.

\begin{array}{l} 2 \sin^{2} θ - 1 = 0 \\ 2 \sin^{2} θ = 1 \\ \sin^{2} θ = \frac{1}{2} \\ \sqrt{\sin^{2} θ} = \pm \sqrt{\frac{1}{2}} \\ \sin θ = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} \\ θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4} \end{array}

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Solving a trigonometric equation involving cosecant

Solve the following equation exactly: $\csc θ = - 2, 0 \leq θ < 4 π .$

We want all values of $θ$ for which $\csc θ = - 2$ over the interval $0 \leq θ < 4 π .$

\begin{array}{l} \csc θ = - 2 \\ \frac{1}{\sin θ} = - 2 \\ \sin θ = - \frac{1}{2} \\ θ = \frac{7 π}{6}, \frac{11 π}{6}, \frac{19 π}{6}, \frac{23 π}{6} \end{array}

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Solving an equation involving tangent

Solve the equation exactly: $\tan (θ - \frac{π}{2}) = 1, 0 \leq θ < 2 π .$

Recall that the tangent function has a period of $π .$ On the interval $[0, π),$ and at the angle of $\frac{π}{4},$ the tangent has a value of 1. However, the angle we want is $(θ - \frac{π}{2}) .$ Thus, if $\tan (\frac{π}{4}) = 1,$ then

\begin{matrix} θ - \frac{π}{2} = \frac{π}{4} \\ θ = \frac{3 π}{4} \pm k π \end{matrix}

Over the interval $[0, 2 π),$ we have two solutions:

\frac{3 π}{4} and \frac{3 π}{4} + π = \frac{7 π}{4}

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Source: OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6

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