Solving Trigonometric Equations Part 2

Trigonometry Tutorial

Solving Trigonometric Equations Part 2

Intro

This is part 2 of 2 on Solving Trigonometric Equations (see Solving Trigonometric Equations Part 1 for problems 1 to 5). In this tutorial I will solve several different trigonometric equations using different techniques such as factoring, substitution and converting the equation to all one trigonometric function using identities. Solving trigonometric equations involves knowledge of several mathematical concepts such as factoring techniques we learn in an algebra class, trigonometric identities and functions and common angle values. Furthermore knowledge of inverse trigonometric functions, recognizing equation forms such as a quadratic equation and creativity are also helpful.

A free online Trigonometry book which I used as a reference can be obtained by going to www.stitz-zeager.com.

Sample Problem

Find all of the exact solutions of the equations and then list those solutions which lie in the interval [0, 2\pi).

6) \quad \sin(2x) = \tan(x)
7) \quad \cot^4(x) = 4\csc^2(x) - 7
8)\quad 4\sin^3(x) - 9\sin^2(x) = \sin(x) - 6
9) \quad \sqrt{3}\sin(3x) - \cos(3x) = \sqrt{3}
10) \quad 2\tan(x) = 1 - \tan^2(x)

Solution

6) \quad \sin(2x) = \cos(x)

To solve this equation we need to use a double angle identity. Then move all terms to one side of the equal sign, factor, set each factor equal to zero and solve for x. You may be tempted to divide each side by \cos(x) after using the identity but don’t do it because you will lose solutions.

The double angle identity that we need is \sin(2\theta) = 2\sin(\theta)\cos(\theta).

    \begin{equation*} 	\begin{align} 	&2\sin(x)\cos(x) = \cos(x)\\ 	&2\sin(x)\cos(x) - \cos(x) = 0\\ 	&\cos(x) \left(2\sin(x) - 1 \right) = 0 	\end{align} \end{equation*}

    \begin{equation*}      \begin{align}           \cos(x) &= 0 \qquad &   2\sin(x) - 1 &= 0 \\           x &= \frac{\pi}{2} + {\pi}k \qquad &   \sin(x) &= 1 \\           x &= \frac{\pi + 2{\pi}k}{2} \qquad &  \sin(x) &= \frac{1}{2}  \\           &{ } \qquad & x &= {\pi}k      \end{align} \end{equation*}

All exact solutions is x = \left\{\frac{\pi + 2{\pi}k}{2}, \thickspace {\pi}k \right\}

The solutions in [0, 2{\pi}) is x = \left\{0, \frac{\pi}{2}, \frac{3\pi}{2}, \pi \right\}
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7) \quad \cot^4(x) = 4\csc^2(x) - 7

To solve this equation we use a Pythagorean identity. Then realize that the new equation has the form of a quadratic equation which then can be solved for using substitution, factoring directly or the quadratic formula.

The Pythagorean indentity we use is 1 + \cot^2(x) = \csc^2(x).

    \begin{equation*} 	\begin{align} 		&\cot^4(x) = 4\csc^2(x) - 7 \\ 		&\cot^4(x) = 4(1 + \cot^2(x)) - 7 \\ 		&\cot^4(x) = 4 + 4\cot^2(x) -7 \\ 		&\cot^4(x) = -3 + 4\cot^2(x) \\                 &\cot^4(x) -4\cot^2(x) + 3 = 0 \\                 &(\cot^2(x) - 1)(\cot^2(x) -3) = 0 \\ 	\end{align} \end{equation*}

    \begin{equation*} 	\begin{align} 		&\cot^2(x) - 1 = 0 \quad & \quad   &\cot^2(x) - 3 = 0 \\ 		&\cot^2(x) = 1 \quad     & \quad   &\cot^2(x) = 3 \\ 		&\cot(x) = \pm 1 \quad   & \quad   &\cot(x) = \pm \sqrt{3} \\ { } \\[20pt] { }         \end{align} \end{equation*}

    \begin{equation*} 	\begin{align} &x = \left\{ \frac{\pi}{4} + {\pi}k, \frac{3\pi}{4} + {\pi}k, \frac{\pi}{6} + {\pi}k, \frac{5\pi}{6} + {\pi}k \right\} \\ { } \\ &\textnormal{All exact solutions is} \thickspace x = \left\{ \frac{\pi + 4{\pi}k}{4}, \frac{3\pi + 4{\pi}k}{4}, \frac{\pi + 6{\pi}k}{6}, \frac{5\pi + 6{\pi}k}{6} \right\} \\ { }\\ &\textnormal{The solutions in $[0, 2{\pi})$ is} \thickspace x = \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{7\pi}{6}, \frac{11\pi}{6} \right\}           \end{align} \end{equation*}

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8)\quad 4\sin^3(x) - 9\sin^2(x) = \sin(x) - 6

To solve this equation move all terms to the left of the equal sign and recognize that the form of the new equation is that of a polynomial. We then let w = \sin(x) and solve for w using the rational zeros theorem. Now replace w with \sin(x) and solve for x.

    \begin{equation*} 	\begin{align} 		&4\sin^3(x) - 9\din^2(x) - \sin(x) + 6 = 0 \\ 		&Let \thickspace w = \sin(x) \\ 		&4w^3 - 9w^2 - w + 6 = 0 \\ 	\end{align} \end{equation*}

We know form the Rational Zeros Theorem that if a polynomial has a rational root(s) then it will be among a list of possible rational zeros formed by taking each factor of the constant divided by each factor of the leading coefficient attaching a plus or minus to the quotient. Note that we do not repeat the same root twice in our final list.

Factors of the constant 6 is {1, 2, 3, 6}.

Factors of the leading coefficient 4 is {1, 2, 4}.

    \begin{equation*} 	\begin{align} & \textnormal{Our list of rational zeros is} \thickspace \pm \frac{1}{1}, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{2}{1}, \pm \frac{2}{2}, \pm \frac{2}{4}, \pm \frac{3}{1}, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{6}{1}, \pm \frac{6}{2}, \pm \frac{6}{4} \\ { } \\ & \textnormal{Simplifying and eliminating repeats we have} \quad w = \left\{ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm 6 \right\} \\ { } \\ &\textnormal{Testing the roots out by using direct substitution or synthetic division we find that} \\ & w =\left\{ -\frac{3}{4}, 1, 2 \right\}. \\ { } \\ % & \sin(x) = -\frac{3}{4} \to x = \arcsin \left( -\frac{3}{4} \right) + 2{\pi}k \\ & \sin(x) = -\frac{3}{4}, \tickspace \textnormal{See below to solve.} \\ { } \\ & \sin(x) = 1 \to x = \left\{ \frac{\pi}{2} + 2{\pi}k \right\} \\ { } \\ & \sin(x) = 2 \to \textnormal{x DNE (does not exist)} 	\end{align} \end{equation*}

One way to solve \sin(x) = -\frac{3}{4} is to use the Reference Angle Theorem. First make the right side positive if necessary and change x to \alpha, the reference angle. The reference angle is the acute angle made with the x-axis and will always be a positive first quadrant angle provided that the angle is not a quadrantal angle using this method. A quadrantal angle is an angle whose terminal side lies on an axis, e.g. \pi. Second determine what quadrants x lies in and draw the angles in on an xy-coordinate system. Third from the drawing determine x.

\sin(\alpha) = \frac{3}{4} \to \alpha = \arcsin\left( \frac{3}{4} \right)

Sin is negative in quadrants 3 and 4.

Rendered by QuickLaTeX.com

From the drawing we see that the angles are

    \begin{equation*} 	\begin{align} & \theta_3 = \pi + \alpha \to x = \pi + \arcsin\left( \frac{3}{4} \right) + 2{\pi}k \\ { } \\ & \theta_4 = 2\pi - \alpha \to x = 2\pi - \arcsin\left( \frac{3}{4} \right) + 2{\pi}k \\ { } \\ & \textnormal{All exact soloutions are} \thickspace x = \left\{\pi + \arcsin\left( \frac{3}{4} \right) + 2{\pi}k, 2\pi - \arcsin\left( \frac{3}{4} \right) + 2{\pi}k \right\} \\ { } \\ & \textnormal{The solutions in $[0, 2{\pi})$ is} \\  & x = \left\{ \pi + \arcsin\left( \frac{3}{4} \right), 2\pi - \arcsin\left( \frac{3}{4} \right) \right\} 	\end{align} \end{equation*}

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9) \quad \sqrt{3}\sin(3x) - \cos(3x) = \sqrt{3}

To solve this equation we could solve as in problem 3 but I will show two alternate methods. The first method involves isolating one trig function then squaring each side. Caution on using this method! Remember that whenever we square each side of an equation we may introduce extraneous solutions, solutions that are valid for the squared equation but not for the original equation. After solving for x remember to check the validity of the solutions.

Method 1:

    \begin{equation*}      \begin{align}           & \sqrt{3}\sin(3x) - \cos(3x) = \sqrt(3) \\           & \sqrt{3}\sin(3x) = \sqrt{3} + \cos(3x) \\           & \textit{Square each side} \\           & \3\sin^2(3x) = 3 + 2\sqrt{3}\cos(3x) + \cos^2(3x) \\           & \textit{Use the identity} \\           & \sin^2(\theta) + \cos^2(\theta) = 1 \\           & 3(1 - \cos^2(3x)) = 3 + 2\sqrt{3}\cos(3x) + \cos^2(3x) \\           & 3 - 3\cos^2(3x)) = 3 + 2\sqrt{3}\cos(3x) + \cos^2(3x) \\           & 0 = 2\sqrt{3}\cos(3x) + 4\cos^2(3x) \\           & 0 = 2\cos(3x)(\sqrt{3} + 2\cos(3x)) \\           & 0 = \cos(3x)(\sqrt{3} + 2\cos(3x)) \\           &{ } \\           & \textit{Set each factor equal to zero and solve for x} \\           & \cos(3x) = 0 \\           & 3x = \frac{\pi}{2} + {\pi}k \\           & 3x = \frac{\pi + 2{\pi}k}{2}  \\           & x = \frac{\pi + 2{\pi}k}{6} \\           & { } \\           & 2\cos(3x) + \sqrt{3} = 0 \to 2\cos(3x) = -\sqrt{3} \to \cos(3x) = -\frac{\sqrt{3}}{2} \\           & 3x = \frac{5\pi}{6} + 2{\pi}k & 3x = \frac{7\pi}{6} + 2{\pi}k \\           & x = \frac{5\pi + 12{\pi}k}{18} & x = \frac{7\pi + 12{\pi}k}{18}      \end{align} \end{equation*}

A list of possible solutions is

x = \left\{ \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}, \frac{5\pi}{18}, \frac{7\pi}{18}, \frac{17\pi}{18}, \frac{19\pi}{18}, \frac{29\pi}{18}, \frac{31\pi}{18} \right\}

Checking the possible solutions in the original equation we find the ones that work are

x = \left\{ \frac{\pi}{6},  \frac{5\pi}{6},  \frac{3\pi}{2},  \frac{5\pi}{18},  \frac{17\pi}{18},  \frac{29\pi}{18} \right\}

To find all exact solutions we use the arithmetic sequence formula a_k = a + d(k - 1), where a is the first term and d is the common difference.

Consider the sequence x = \left\{ \frac{\pi}{6},  \frac{5\pi}{6},  \frac{3\pi}{2}=\frac{9\pi}{6}, \ldots  \right\}. We see that the first term is \frac{\pi}{6} and the common difference is \frac{4\pi}{6}. Therefore all exact solutions for this part is

    \begin{equation*}      \begin{align}           a_{k} &=  \frac{\pi}{6} +\frac{4\pi}{6}(k - 1) \\                 & = \frac{\pi}{6}\big(1 + 4(k - 1)\big) \\                 & = \frac{\pi}{6}(1 + 4k - 4) \\                 & = \frac{\pi}{6}( 4k - 3) \\                 & { } \\                 & \textit{For the sequence} \thickspace x = \left\{ \frac{5\pi}{18}, \frac{17\pi}{18},  \frac{29\pi}{18} \right\} \\           b_{k} &=  \frac{\pi}{18}(12k - 7) \\      \end{align} \end{equation*}

All exact solutions is x = \left\{ \frac{\pi}{6}( 4k - 3), \frac{\pi}{18}(12k - 7) \right\}

The solutions in [0, 2\pi) is x = \left\{ \frac{\pi}{6},  \frac{5\pi}{6},  \frac{3\pi}{2},  \frac{5\pi}{18},  \frac{17\pi}{18},  \frac{29\pi}{18} \right\}

When solving trigonometric equations I do not recommend this method.

Method 2.

9) \quad \sqrt{3}\sin(3x) - \cos(3x) = \sqrt{3}

Use to coefficients of the trig functions as the legs of a right triangle drawing it in standard position in the correct quadrant. Then use Pythagorean’s theorem to find the hypotenuse. Determine \theta, the angle from the positive x-axis to the terminal side of the angle.

The coefficients of the sine and cosine functions are \sqrt{3} and -1 respectively. Creating a right triangle letting x be the \sqrt{3} and y be -1 (we could have chosen x to be -1 and y to be \sqrt{3}) we have

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From the triangle we see that
r = \sqrt{(\sqrt{3})^2 + (-1)^2} = 2
\alpha = \arctan(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}

Remember that the reference angle \alpha is always the positive acute angle with respect to the x-axis. We have two choices for \theta, \thickspace -\frac{\pi}{6} or \frac{11\pi}{6}. Let us pick -\frac{\pi}{6}.

Using the triangle and the definition of sine and cosine we see that

    \begin{equation*}      \begin{align}          & \sin(\theta) = -\frac{1}{2} \to 2\sin(\theta) = -1 \\          &\cos(\theta) = \frac{\sqrt{3}}{2} \to 2\cos(\theta) = \sqrt{3} \\          &\textit{Replacing the coefficients with their equivalents we have} \\          &2\cos(\theta)\sin(3x) + 2\sin(\theta)\cos(3x) = \sqrt{3} \to 2\sin(3x + \theta) = \sqrt{3} \\          &\sin(3x - \frac{\pi}{6}) = \frac{\sqrt{3}}{2} \\          3x - \frac{\pi}{6} &= \frac{\pi}{3} + 2{\pi}k & 3x - \frac{\pi}{6} &= \frac{2\pi}{3} + 2{\pi}k \\          3x  &= \frac{\pi}{3} + \frac{\pi}{6} + 2{\pi}k & 3x &= \frac{2\pi}{3} + \frac{\pi}{6} + 2{\pi}k \\          3x  &= \frac{\pi}{2} + 2{\pi}k & 3x &= \frac{5\pi}{6} + 2{\pi}k \\          3x &= \frac{\pi + 4{\pi}k}{2} & 3x &= \frac{5\pi + 12{\pi}k}{6} \\          x &= \frac{\pi + 4{\pi}k}{6} & x &= \frac{5\pi + 12{\pi}k}{18}       \end{align} \end{equation*}

All exact solutions is x = \left\{ \frac{\pi}{6}( 1 + 4k), \frac{\pi}{18}(5 + 12k) \right\}

The solutions in [0, 2\pi) is x = \left\{ \frac{\pi}{6},  \frac{5\pi}{6},  \frac{3\pi}{2},  \frac{5\pi}{18},  \frac{17\pi}{18},  \frac{29\pi}{18} \right\}

Note that although these sequence formulas are different from the previous ones they generate the same numerical values.
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10) \quad 2\tan(x) = 1 - \tan^2(x)

I have already solved this equation, number 5, in Solving Trigonometric Equations Part 1 and will show an alternate method. We can solve this equation by recognizing that it is the double angle formula for the tangent whose identity is

\tan\left(2x\right) = \frac{2\tan(x)}{1 - \tan^2(x)}

    \begin{equation*}     \begin{align}           &2\tan(x) = 1 - \tan^2(x) \\           &\textit{Divide each side by $1 - \tan^2(x)$} \\           &\frac{2\tan(x)}{1 - \tan^2(x)} = 1 \\           &\tan(2x) = 1 \to 2x = \frac{\pi}{4} + {\pi}k \to x = \frac{\pi}{8}(1+4k)      \end{align} \end{equation*}

All exact solutions is x = \frac{\pi}{8}(1+4k)

The solutions in [0, 2\pi) is x = \left\{ \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8} \right\}

Although this solution appears to different from the one in problem 5 it is equivalent. Use your calculator to verify that both solutions are indeed equivalent.



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