Equations & Inequalities with Sinx, Cosx and Tanx (College Board AP® Precalculus): Revision Note

Solving simple trigonometric equations

What are simple trigonometric equations?

• A simple trigonometric equation is one of the form

  • $\sin \theta =k,\cos \theta =k,\text{or}\tan \theta =k$

    • where $k$ is a constant

• The goal is to find all values of $\theta$ in a specified solution interval that make the equation true

• Because trigonometric functions are periodic

  • there are usually infinitely many solutions to a trigonometric equation across all real numbers

  • but a question typically restricts the solution interval

    • e.g. $0\le \theta <2\pi$

How can I find an initial solution?

• The first step is to find an initial solution

  • i.e. any one specific value of $\theta$ that satisfies the equation

• For values of $k$ corresponding to special angles (multiples of $\frac{\pi }{6}$ or $\frac{\pi }{4}$)

  • you can use your knowledge of exact trigonometric values from the unit circle

    • E.g. for $\sin \theta =\frac{\sqrt{3}}{2}$

    • recognize that $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$

      • so $\theta =\frac{\pi }{3}$ is an initial solution

• On a calculator part of the exam

  • you can use an inverse trigonometric function on your graphing calculator

    • E.g. for $\sin \theta =0.4$

      • an initial solution is $\theta ={\sin }^{-1}(0.4)\approx 0.4115$

How can I find other solutions from the initial solution?

• Once one solution is known, additional solutions in the interval can be found using the symmetry and periodicity of the trigonometric functions

• For sine equations ($\sin \theta =k$)

  • If ${\theta }_0$ is one solution, then $\pi -{\theta }_0$ is another solution

    • by symmetry of the sine graph about $\theta =\frac{\pi }{2}$ in the unit circle (see diagram below)

  • All other solutions can then be obtained

    • by adding or subtracting integer multiples of $2\pi$ (the period) to the first two solutions

  • Note that subtracting $2\pi$ from $\pi -{\theta }_0$ gives $-\pi -{\theta }_0$

    • If ${\theta }_0$ is in the interval $\left[-\pi ,0\right]$, this automatically gives another solution in$\left[-\pi ,0\right]$

  • Note that adding $2\pi$ to $\pi -{\theta }_0$ gives $3\pi -{\theta }_0$

    • If ${\theta }_0$ is in the interval $\left[\pi ,2\pi \right]$, this automatically gives another solution in$\left[\pi ,2\pi \right]$

• For cosine equations ($\cos \theta =k$)

  • If ${\theta }_0$ is one solution, then $-{\theta }_0$ is another solution

    • by symmetry of the cosine graph about $\theta =0$ in the unit circle (see diagram below)

    • This also corresponds with cosine being an even function

      • i.e. $\cos \left(-\theta \right)=\cos \theta$ in general

  • All other solutions can then be obtained

    • by adding or subtracting integer multiples of $2\pi$ (the period) to the first two solutions

  • Note that adding $2\pi$ to $-{\theta }_0$ gives $2\pi -{\theta }_0$

    • If ${\theta }_0$ is in the interval $\left[0,2\pi \right]$, this automatically gives another solution in$\left[0,2\pi \right]$

• For tangent equations ($\tan \theta =k$):

  • All other solutions differ from the principal solution by integer multiples of $\pi$

    • since tangent has period $\pi$

What is the general process for solving a trigonometric equation in a given interval?

• Start by rearranging the equation (if needed) so that it has the form $\sin \theta =k$, $\cos \theta =k$, or $\tan \theta =k$

  • If the equation involves a trigonometric function appearing more than once (e.g. $2{\cos }^2\theta =\cos \theta$), use algebra (typically factoring) to break it into simpler equations

    • $2{\cos }^2\theta -\cos \theta =0\Rightarrow \cos \theta \left(2\cos \theta -1\right)=0\Rightarrow \cos \theta =0\mathrm{or}\cos \theta =\frac{1}{2}$

• Next find an initial solution

  • using exact values or a calculator

• Then use symmetry and periodicity

  • to find all other solutions in the specified interval

• Finally check that all solutions are within the given interval

  • and discard any that are not

How can the symmetry properties be visualized?

• The unit circle is the most reliable tool for visualizing how many solutions exist and where they are

• For $\sin \theta =k$ (with $-1<k<1$, $k\ne 0$)

  • there are two solutions in any interval of length $2\pi$

  • located symmetrically across the vertical line $\theta =\frac{\pi }{2}$

    • If $\sin \theta =k$, then $\sin \left(\pi -\theta \right)=k$ is also true

• For $\cos \theta =k$ (with $-1<k<1$, $k\ne 0$)

  • there are two solutions in any interval of length $2\pi$

  • located symmetrically across the horizontal axis (i.e. $\theta =0$)

    • If $\cos \theta =k$, then $\cos \left(-\theta \right)=k$ and $\cos \left(2\pi -\theta \right)=k$ are also true

• For $\tan \theta =k$

  • there is one solution in any interval of length $\pi$

  • and all other solutions are integer multiple of $\pi$ apart

    • If $\tan \theta =k$, then it's also true that $\tan \left(\theta -\pi \right)=k$, $\tan \left(\theta +\pi \right)=k$, etc.

How are 'hidden quadratic' trigonometric equations solved?

• Some trigonometric equations are quadratics in disguise

  • They have the form of a quadratic equation

    • but with a trigonometric function in place of the variable

  • For example $2{\cos }^{2}x+7\cos x+3=0$

    • has the same structure as the algebraic equation $2y^2+7y+3=0$

• These equations can be solved using a substitution

• E.g. to solve $2{\cos }^{2}x+7\cos x+3=0$ in the interval $[0,2\pi )$

  • Let $\mathit{y}$ (or another variable) stand for the trigonometric function

    • I.e. $y=\cos x$

    • The equation becomes a standard quadratic in $y$

      • $2y^2+7y+3=0$

  • Solve the quadratic in $y$ by factoring or using the quadratic formula

    • $(2y+1)(y+3)=0$, giving $y=-\frac{1}{2}$ or $y=-3$

  • Substitute the trigonometric function back in

    • $\cos x=-\frac{1}{2}$ or $\cos x=-3$

  • Some solutions may need to be rejected

    • For sine and cosine, only values with $-1\le y\le 1$ are valid

      • any other value of $y$ corresponds to no solution and must be discarded

      • So in this case $\cos x=-3$ must be rejected

    • For tangent, all real values of $y$ are valid

  • $\cos x=-\frac{1}{2}$ has initial solution $x=\frac{2\pi }{3}$

    • which means that, by symmetry, $x=2\pi -\frac{2\pi }{3}=\frac{4\pi }{3}$ is also a solution

  • So the only solutions to $2{\cos }^{2}x+7\cos x+3=0$ in $[0,2\pi )$ are $x=\frac{2\pi }{3}, \frac{4\pi }{3}$

Solving simple trigonometric inequalities

How can I solve a trigonometric inequality?

• A simple trigonometric inequality has the form $\sin \theta <k$, $\cos \theta \ge k$, etc.

  • The solution is an interval (or a union of intervals) within the specified solution interval

• In general

  • Start by solving the corresponding equation

    • i.e. with $=$ instead of the inequality sign

    • This will give you the boundary values of $\theta$

  • Use a graph (or the unit circle) to determine

    • which intervals between (or outside) those boundary values satisfy the inequality

  • Finally, check the endpoints

    • Include them if the inequality is non-strict ($\le$ or $\ge$)

    • Exclude them if the inequality is strict ($<$ or $>$)

How can I solve a system of trigonometric inequalities?

• If a question requires more than one inequality to be satisfied at the same time:

  • Find the solution set for each inequality separately

  • then take the intersection

    • i.e. find the values of $\theta$ that satisfy all of them

• Sketching the solution intervals on a number line can help identify the intersection clearly