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Solving Trigonometric Equations Using Basic Algebra

Substitute in potential solutions or solve using inverse trig functions.

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Solving Trigonometric Equations Using Algebra

As Agent Trigonometry, you are given this clue: \begin{align*}2 \sin x-\sqrt{2}=0\end{align*}. If \begin{align*}0 \le x < 2 \pi\end{align*}, what is/are the value(s) of \begin{align*}x\end{align*}.

Solving Trigonometric Equations

We have already verified trigonometric identities, which are true for every real value of \begin{align*}x\end{align*}. In this concept, we will solve trigonometric equations. An equation is only true for some values of \begin{align*}x\end{align*}.

Let's verify that \begin{align*}\csc x-2=0\end{align*} when \begin{align*}x=\frac{5 \pi}{6}\end{align*}.

Substitute in \begin{align*}x=\frac{5 \pi}{6}\end{align*} to see if the equations holds true.

\begin{align*}\csc \left(\frac{5 \pi}{6}\right)-2&=0 \\ \frac{1}{\sin \left(\frac{5 \pi}{6}\right)}-2&=0 \\ \frac{1}{\frac{1}{2}}-2&=0 \\ 2-2&=0\end{align*}

This is a true statement, so \begin{align*}x=\frac{5 \pi}{6}\end{align*} is a solution to the equation.

Now, let's solve \begin{align*}2 \cos x+1=0\end{align*}.

To solve this equation, we need to isolate \begin{align*}\cos x\end{align*} and then use inverse to find the values of \begin{align*}x\end{align*} when the equation is valid.

\begin{align*}2 \cos x+1&=0 \\ 2 \cos x&=-1 \\ \cos x&=- \frac{1}{2}\end{align*}

So, when is the \begin{align*}\cos x=- \frac{1}{2}\end{align*}? Between \begin{align*}0 \le x< 2 \pi, x=\frac{2 \pi}{3}\end{align*} and \begin{align*}\frac{4 \pi}{3}\end{align*}. But, the trig functions are periodic, so there are more solutions than just these two. You can write the general solutions as \begin{align*}x=\frac{2 \pi}{3} \pm 2 \pi n\end{align*} and \begin{align*}x=\frac{4 \pi}{3} \pm 2 \pi n\end{align*}, where \begin{align*}n\end{align*} is any integer. You can check your answer graphically by graphing \begin{align*}y=\cos x\end{align*} and \begin{align*}y=- \frac{1}{2}\end{align*} on the same set of axes. Where the two lines intersect are the solutions.

Finally, let's solve \begin{align*}5 \tan(x+2)-1=0\end{align*}, where \begin{align*}0 \le x < 2 \pi\end{align*}.

In this problem, we have an interval where we want to find \begin{align*}x\end{align*}. Therefore, at the end of the problem, we will need to add or subtract \begin{align*}\pi\end{align*}, the period of tangent, to find the correct solutions within our interval.

\begin{align*}5 \tan(x+2)-1&=0 \\ 5 \tan(x+2)&=1 \\ \tan(x+2)&=\frac{1}{5}\end{align*}

Using the \begin{align*}\tan^{-1}\end{align*} button on your calculator, we get that \begin{align*}\tan^{-1} \left(\frac{1}{5}\right)=0.1974\end{align*}. Therefore, we have:

\begin{align*}x+2&=0.1974 \\ x&=-1.8026\end{align*}

This answer is not within our interval. To find the solutions in the interval, add \begin{align*}\pi\end{align*} a couple of times until we have found all of the solutions in \begin{align*}[0, 2 \pi]\end{align*}.

\begin{align*}x&=-1.8026+ \pi=1.3390 \\ &=1.3390+ \pi=4.4806\end{align*}

The two solutions are \begin{align*}x = 1.3390\end{align*} and 4.4806.


Example 1

Earlier, you were asked to find the value of x from the equation \begin{align*}2 \sin x-\sqrt{2}=0\end{align*}

To solve this equation, we need to isolate \begin{align*}\sin x\end{align*} and then use inverse to find the values of \begin{align*}x\end{align*} when the equation is valid.

\begin{align*}2 \sin x -\sqrt{2}=0 \\ 2 \sin x&=\sqrt{2} \\ \sin x&=\frac{\sqrt{2}}{2}\end{align*}

So now we need to find the values of \begin{align*}x\end{align*} for which \begin{align*}\sin x =\frac{\sqrt{2}}{2}\end{align*}. We know from the special triangles that this value of sine holds true for a \begin{align*}45^\circ\end{align*} angle, but is that the only value of \begin{align*}x\end{align*} for which it is true?

We are told that \begin{align*}0 \le x < 2 \pi\end{align*}. Recall that the sine is positive in both the first and second quadrants, so \begin{align*}\sin x =\frac{\sqrt{2}}{2}\end{align*} when \begin{align*}x\end{align*} also is \begin{align*}135^\circ\end{align*}.

Example 2

Determine if \begin{align*}x=\frac{\pi}{3}\end{align*} is a solution for \begin{align*}2 \sin x=\sqrt{3}\end{align*}.

\begin{align*}2 \sin \frac{\pi}{3}= \sqrt{3} \rightarrow 2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}\end{align*} Yes, \begin{align*}x=\frac{\pi}{3}\end{align*} is a solution.

Example 3

Solve the following trig equation in the interval \begin{align*}0 \le x< 2 \pi\end{align*}.

\begin{align*}3 \cos^2x-5=0\end{align*}

\begin{align*}9 \cos^2x-5&=0 \\ 9 \cos^2x&=5 \\ \cos^2x&=\frac{5}{9} \\ \cos x&=\pm \frac{\sqrt{5}}{3}\end{align*}

The \begin{align*}\cos x=\frac{\sqrt{5}}{3}\end{align*} at \begin{align*}x=0.243\end{align*} rad (use your graphing calculator). To find the other value where cosine is positive, subtract 0.243 from \begin{align*}2 \pi\end{align*}, \begin{align*}x=2 \pi -0.243=6.037\end{align*} rad.

The \begin{align*}\cos x=- \frac{\sqrt{5}}{3}\end{align*} at \begin{align*}x=2.412\end{align*} rad, which is in the \begin{align*}2^{nd}\end{align*} quadrant. To find the other value where cosine is negative (the \begin{align*}3^{rd}\end{align*} quadrant), use the reference angle, 0.243, and add it to \begin{align*}\pi\end{align*}. \begin{align*}x= \pi+0.243=3.383\end{align*} rad.

Example 4

Solve the following trig equation in the interval \begin{align*}0 \le x< 2 \pi\end{align*}.

\begin{align*}3 \sec(x-1)+2=0\end{align*}

Here, we will find the solution within the given range, \begin{align*}0 \le x< 2 \pi\end{align*}.

\begin{align*}3 \sec(x-1)+2&=0 \\ 3 \sec(x-1)&=-2 \\ \sec(x-1)&=- \frac{2}{3} \\ \cos(x-1)&=- \frac{3}{2}\end{align*}

At this point, we can stop. The range of the cosine function is from 1 to -1. \begin{align*}- \frac{3}{2}\end{align*} is outside of this range, so there is no solution to this equation.


Determine if the following values for \begin{align*}x\end{align*}. are solutions to the equation \begin{align*}5+6 \csc x=17\end{align*}.

  1. \begin{align*}x=- \frac{7 \pi}{6}\end{align*}
  2. \begin{align*}x=\frac{11 \pi}{6}\end{align*}
  3. \begin{align*}x=\frac{5 \pi}{6}\end{align*}

Solve the following trigonometric equations. If no solutions exist, write no solution.

  1. \begin{align*}1- \cos x=0\end{align*}
  2. \begin{align*}3 \tan x - \sqrt{3}=0\end{align*}
  3. \begin{align*}4 \cos x=2 \cos x+1\end{align*}
  4. \begin{align*}5 \sin x-2=2 \sin x+4\end{align*}
  5. \begin{align*}\sec x-4=- \sec x\end{align*}
  6. \begin{align*}\tan^2(x-2)=3\end{align*}

Sole the following trigonometric equations within the interval \begin{align*}0 \le x < 2 \pi\end{align*}. If no solutions exist, write no solution.

  1. \begin{align*}\cos x=\sin x\end{align*}
  2. \begin{align*}- \sqrt{3} \csc x=2\end{align*}
  3. \begin{align*}6 \sin(x-2)=14\end{align*}
  4. \begin{align*}7 \cos x -4=1\end{align*}
  5. \begin{align*}5+4 \cot^2x=17\end{align*}
  6. \begin{align*}2 \sin^2x-7=-6\end{align*}

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.10. 

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