Solving Trigonometric Equations Part 1

Trigonometry Tutorial

Solving Trigonometric Equations Part 1

Intro

This tutorial is Part 1 of 2 on Solving Trigonometric Equations. 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).

1) \quad \sqrt{2}\cos(2x) + 1 = 0
2) \quad \cos(3x) = \cos(5x)
3)\quad \sqrt{3}\sin(2x) + \cos(2x) = 1
4) \quad 2\sec^2(x) = 3 - \tan(x)
5) \quad 2\tan(x) = 1 - \tan^2(x)

Solution

1) \quad \sqrt{2}\cos(2x) + 1 = 0

Since we have a single trig function our objective is to isolate the trig function using algebra then solve for x.

Step 1) Isolate the trig function using algebra.

Subtract 1 from each side;

\sqrt{2} \cos(2x) = -1

Divide each side by \sqrt{2}

\cos(2x) = \frac{-1} {\sqrt{2}}

Step 2) Replace the argument of 2x with a general letter such as u. Then solve for u in the interval [0, 2\pi] adding 2 {\pi} k, k an integer, to each value. Remember that the trig functions sin, cos, csc and sec are periodic in 2\pi.

Solve \cos(u) = \frac{-1} {\sqrt{2}}.

The angles at which cos is \frac{-1} {\sqrt{2}} is \frac{3 \pi}{4} or \frac{5 \pi}{4}. These values can be obtained from a figure of the unit circle with common angle values or from memory. Therefore

u = \left\{ \frac{3 \pi}{4} + 2 {\pi} k, \quad \frac{5 \pi}{4} + 2 {\pi} k \right\}

Step 3) Replace u with the original argument of 2x and solve for x.

2x = \frac{3 \pi}{4} + 2 {\pi} k \quad or \quad 2x = \frac{5 \pi}{4} + 2 {\pi} k \right}

8x = 3 \pi + 8 {\pi} k \quad or \quad 8x = 5 \pi + 8 {\pi} k

x = \frac{3 \pi + 8 {\pi} k}{8} \quad or \quad x = \frac{5 \pi + 8 {\pi} k}{8}

All exact solutions is x = \left\{ \frac{3 \pi + 8 {\pi} k}{8}, \thickspace \frac{5 \pi + 8 {\pi} k}{8} \right\}

You may be wondering why I put the solution for all values over one denominator. I did this to make the calculations easier to find all values of x in the interval [0, 2 \pi).

Substituting 0, 1, 2, 3, … in for k until x is greater than 2 \pi gives

The solutions in [0, 2{\pi}) is x = \left\{ \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{11 \pi}{8}, \frac{13 \pi}{8} \right\}
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2) \quad \cos(3x) = \cos(5x)

\cos(3x) - \cos(5x) = 0 \qquad (1)

To solve this equation we must recognize that the left hand side of (1) resembles a sum to product identity. Our objective is to write the left hand side as the product of two sine functions using the sum to product formulas then solve for x.

The sum to product identity that we need is

\cos(\alpha) - \cos(\beta) = -2\sin\left(\frac{\alpha + \beta}{2}\right)\sin\left(\frac{\alpha - \beta}{2}\right) \qquad (2)

Comparing the left sides of (1) and (2) we see that \alpha = 3x and \beta = 5x}. Substituting these values into (2) and using the identity \sin(-\theta) = -\sin(\theta) we have

    \begin{equation*}      \begin{align}           \cos(3x) - \cos(5x) &= -2\sin\left(\frac{3x + 5x}{2}\right)\sin\left(\frac{3x - 5x}{2}\right) \\                               &= -2\sin(4x)\sin(-x) \\                               &= 2\sin(4x)\sin(x), \thickspace \textnormal{therefore} 	\end{align} \end{equation*}

2\sin(4x)\sin(x) = 0 \to \sin(4x)=0 \quad or \quad \sin(x)=0

The angles at which sine is 0 is 0 or \pi. Since these two values are \pi apart we only need to add {\pi}k to the smaller value.

4x = 0 + {\pi}k \to 4x = {\pi}k \to x = \frac{{\pi}k}{4} \quad or \quad x = {\pi}k

Note that x = {\pi}k is not needed because x = \frac{{\pi}k}{4} generates the same values when k is a multiple of 4. So my solution for all values is

All exact solutions is x = \left\{ \frac{{\pi}k}{4} \right\}

Substituting 0, 1, 2, 3, … in for k until x is greater than 2 \pi gives

The solutions in [0, 2{\pi}) is x = \left\{ 0, \frac{\pi}{4}, \frac{\pi}{2}. \frac{3\pi}{4}, {\pi}, \frac{5\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4} \right\}
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3)\quad \sqrt{3}\sin(2x) + \cos(2x) = 1 \qquad (1)

The objective here is to write the expression on the left side of the equation as a sine or cosine function of the form A\cos({\omega}x + \phi) or A\sin({\omega}x + \phi). Note that this method can only be applied if the arguments are the same. The left side resembles part of the sum or difference formulas for sine or cosine. I pick cosine.

A\cos({\omega}x + \phi) = A\cos({\omega}x)\cos(\phi) - A\sin({\omega}x)\sin(\phi) \qquad (2)

Set the left side of (1) equal to the right side of (2) we have

\sqrt{3}\sin(2x) + \cos(2x) = A\cos({\omega}x)\cos(\phi) - A\sin({\omega}x)\sin(\phi)

By comparing each side and equating coefficients we see that

\omega = 2,\quad A\cos(\phi) = 1, \quad -A\sin(\phi) = \sqrt{3} \longrightarrow A\sin(\phi) = -\sqrt{3}

To solve for A use the identity \cos^2(\phi) + \sin^2(\phi) = 1

Multiply each side by A^2.

A^2\cos^2(\phi) + A^2\sin^2(\phi) = A^2 \qquad (3)

Substituting A\cos(\phi) = 1 \quad and \quad A\sin(\phi) = -\sqrt{3} into (3) gives

(1)^2 + (-\sqrt{3})^2 = A^2 \longrightarrow 4 = A^2 \longrightarrow A = {\pm} 2

I will pick A = 2. You could use A = -2. Now solve for \phi.

2\cos(\phi) = 1 \quad and \quad 2\sin(\phi) = -\sqrt{3}

\cos(\phi) = \frac{1}{2} \quad and \quad \sin(\phi) = \frac{-\sqrt{3}}{2}

The value of \phi that satisfies both equations is \frac{5\pi}{3} \thickspace or \thickspace \frac{-\pi}{3}. Either angle will work. I pick \phi = \frac{-\pi}{3}. Substituting A and \phi into A\cos({\omega}x + \phi}) we have

\sqrt{3}\sin(2x) + \cos(2x) = 2\cos(2x - \frac{\pi}{3}) \longrightarrow 2\cos(2x - \frac{\pi}{3}) = 1

Now solve for x.

\cos(2x - \frac{\pi}{3}) = \frac{1}{2}

The angles for which cosine is \frac{1}{2} is \frac{\pi}{3} \thickspace or \thickspace \frac{5\pi}{3}.

2x - \frac{\pi}{3} = \frac{\pi}{3} + 2{\pi}k\quad or \quad 2x - \frac{\pi}{3} = \frac{5\pi}{3} + 2{\pi}k

2x - \frac{\pi}{3} = \frac{\pi}{3} + 2{\pi}k \\ 6x - \pi = \pi + 6{\pi}k \\ 6x = 2\pi + 6{\pi}k \\ x = \frac{2\pi +6{\pi}k}{6} \\ x = \frac{2(\pi +3{\pi}k)}{6} \\ x = \frac{\pi +3{\pi}k}{3} \qquad (4)

2x - \frac{\pi}{3} = \frac{5\pi}{3} + 2{\pi}k \\ 6x - \pi = 5\pi + 6{\pi}k \\ 6x = 6\pi + 6{\pi}k \\ x = \pi +{\pi}k \qquad (5)

All exact solutions is x = \left\{ \frac{\pi +3{\pi}k}{3}, \thickspace \pi +{\pi}k \right\}

Substituting 0, 1, 2, 3, … in for k in (4) and -1, 0, 1, 2, 3, … in for k in (5) until x is greater than 2 \pi gives

The solutions in [0, 2{\pi}) is x= \left\{ 0, \frac{\pi}{3}, {\pi}, \frac{4\pi}{3} \right\}

Can you see why I started off with k = -1 in (5)? Remember we want to list those solutions which are in the interval [0, 2{\pi}). Do not always assume that k starts at zero.
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4) \quad 2\sec^2(x) = 3 - \tan(x)

To solve this equation we need to move all terms to one side of the equal sign and get the same trig function using a Pythagorea 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 identity that we need is 1 + \tan^2(x) = \sec^2(x).

2\sec^2(x) + \tan(x) - 3 = 0 \\ 2(1 + \tan^2(x)) + \tan(x) -3 = 0 \\ 2 + 2\tan^2(x) + tan(x) - 3 = 0 \\ 2\tan^2(x) + \tan(x) -1 = 0 \quad (1)

Note that we can factor (1) directly by considering tan(x) as the variable so

2\tan^2(x) + \tan(x) -1 = 0 \to (2\tan(x) - 1)(\tan(x) + 1) =0

If you cannot see how I factored you can use the substitution method that I will demonstrate next. Remember that tan and cot are periodic every {\pi}.

Solving using the substitution method

Let \thickspace w = \tan(x) \to w^2 = \tan^2(x) \\ 2w^2 + w -1 = 0 \to (2w - 1)(w + 1) = 0 \\ 2w - 1 = 0 \to 2w = 1 \to w = \frac{1}{2} \quad or \quad w + 1 = 0 \to w = -1

Replacing w with \tan(x)

\tan(x) = \frac{1}{2} \to x = \arctan(\frac{1}{2}) + {\pi}k \quad or \quad \tan(x) = -1 \to x = \frac{-\pi}{4} + {\pi}k = \frac{-\pi + 4{\pi}k}{4}

Note that there is no common angle for \tan(x) = \frac{1}{2} so we use the arctan function, aka inverse trig functions. Remember that the inverse functions are defined over specific intervals and only gives one angle. We know that tangent is positive in quadrants 1 and 3; however, the arctan function gives only the first quadrant angle. Remember that if the two angles are \pi apart we only need to add {\pi}k to the smaller angle.

All exact solutions is x = \left\{ \arctan(\frac{1}{2}) + {\pi}k, \thickspace \frac{-\pi + 4{\pi}k}{4} \right\}

The solutions in [0, 2{\pi}) is x = \left\{ \arctan(\frac{1}{2}), \arctan(\frac{1}{2}) + \pi, \frac{3\pi}{4}, \frac{7\pi}{4} \right\}
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5) \quad 2\tan(x) = 1 - \tan^2(x)

To solve this equation we move all terms to the left side of the equal sign and realize that the new equation has the form of a quadratic equation which can be solved by factoring if possible or by using the quadratic formula.

\tan^2(x) + 2\tan(x) -1 = 0 \qquad (1)

Since (1) does not factor use the quadratic formula treating \tan(x) as the variable

    \begin{equation*} \left. \begin{aligned} \tan(x) &= \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-1)}}{2(1)}\\               &= \frac{-2 \pm \sqrt{8}}{2}}\\               &= \frac{-2 \pm 2\sqrt{2}}{2}}\\               &= \frac{2(-1 \pm \sqrt{2}}{2}}\\               &= -1 \pm \sqrt{2} \end{aligned} \end{equation*}

\tan(x) = -1 - \sqrt{2} \to x = \arctan(-1 - \sqrt{2}) + {\pi}k \quad or \quad \tan(x) = -1 + \sqrt{2} \to x = \arctan(-1 + \sqrt{2}) + {\pi}k

All exact solutions is x = \left\{ \arctan(-1 - \sqrt{2}) + {\pi}k, \thickspace \arctan(-1 + \sqrt{2}) + {\pi}k \right\}

The solutions in [0, 2{\pi}) is x = \left\{ \arctan(-1 - \sqrt{2}) + \pi, \thickspace \arctan(-1 - \sqrt{2}) + 2\pi, \thickspace \arctan(-1 + \sqrt{2}), \thickspace \arctan(-1 + \sqrt{2}) + \pi \}



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College Math, Statistics And Trigonometry
I am a graduate of Cleveland State University. After earning my Bachelors of Electrical Engineering I went on to earn a Master’s of Science in Electrical Engineering. Also I am a graduate of Lorain County Community College (LCCC) earning an Associates of Science Pre-Professional Engineering and ...
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