Solving Trigonometric Equations Using Multiple Strategies

1. Solving Trigonometric Equations That Are Quadratic in Form

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To step through solving the equation on the interval take a look at this video.

EXAMPLE

Consider the equation on the interval

This equation is quadratic in form since the expression contains a square term and a linear term in cosine, and a constant term.

	Let then substitute.
	Factor.
	Set and solve.
	Set and solve.

Next, back-substitute so we can solve the equation for x.

	Back-substitute into the equation
	Using the unit circle, these are the angles that correspond to
	Back-substitute into the equation
	From the unit circle,

Thus, the solutions are

The graph of is shown below. Note the locations of the x-intercepts on the interval

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Consider the equation

Find all exact solutions on the interval [0, 2π).

	Original equation.
	Factor from the expression.
	Set each factor equal to 0.
	From the unit circle, when and when
	In the second equation, isolate to one side. Then, consult the unit circle to find the solutions.

There are four solutions to this equation:

2. Solving Trigonometric Equations Using Fundamental Identities

There are instances where the equation can’t be solved easily as written, so an identity is used to rewrite.

EXAMPLE

Consider the equation on the interval

Notice that there are two different trigonometric functions in this equation. Using an identity, we can rewrite the equation and solve it.

	This is the original equation.
	Use the Pythagorean identity
	Combine like terms.
	Factor out

Next, set each factor equal to 0 and solve.

	Set the first factor equal to 0.
	when is an odd multiple of
	Set the second factor equal to 0.
	Add to both sides and write on the left side.
	From the unit circle,

Thus, the solutions are

try it

Consider the equation

Rewrite the equation using a Pythagorean identity, then isolate the trigonometric expression to one side.

If using then the equation becomes:

	Original equation.
	Replace with
	Add to both sides.
	Divide both sides by 2.

If using then the equation becomes:

	Original equation.
	Replace with
	Add to both sides.
	Divide both sides by 2.

Either of these equations can be used to solve for which is in the next part.

Find all exact solutions on the interval [0, 2π).

Taking the equation solve for

	This is the starting point.
	Apply the square root principle, then rationalize the denominator.

This leads to two equations:

which has solutions

which has solutions

Therefore, the solutions to the equation are

EXAMPLE

Consider the equation

This equation contains two sine terms, but they have different angles. We will need to use a double angle identity to rewrite the expression on the left side.

	This is the original equation.
	Factor.

Now, set each factor equal to 0 and solve.

	Set the first factor equal to 0.
	when x is an integer multiple of
	Set the second factor equal to 0.
	Subtract 3 from both sides, then divide by 2.
No solution	Since is not in the range of the cosine function, there is no solution to this equation.

Thus, the solutions to the equation are

To get a graphical perspective of the solutions, graph the function on the interval and note that its x-intercepts are and

try it

Consider the equation

Using an appropriate identity, write an equation set to zero so that the expression on the left side is in terms of cos θ.

Choose since there are only cosine terms (the others contain sine terms).

Then, the equation becomes or in descending order,

Find all exact solutions on the interval [0, 2π).

Notice that this equation appears quadratic. Let

Then, the equation becomes

This factors nicely:

Set each factor equal to 0 and solve:


Now, replace u with

means which has solutions

means which has the solution

Therefore, the solutions to the equation are

This next example is quite different from the others since the above strategies do not apply. How will we solve it?

EXAMPLE

Consider the equation Find all solutions on the interval

There is no identity that can be used to replace with an expression involving or vice versa.

One thing to remember is that and are related through a Pythagorean identity, so it may be worth a try to square both sides of the equation.

	This is the original equation.
	Square both sides.
	Expand the right side.
	Replace with
	Add to both sides, and subtract 1 from both sides. Then, write the expression on the left side.
	Factor out

Now, set each factor equal to 0 and solve.

	Set the first factor equal to 0.
	Divide both sides by 2.
	when x is an odd multiple of
	Set the second factor equal to 0.
	Subtract 1 from both sides.
	This is the solution.

Therefore, we have three potential solutions:

Now, substitute each value into the original equation

Potential Solution	Left Side of Equation	Right Side of Equation	Does It Check?
			Yes
			Yes
			No

Therefore, the solutions are

To see the graphical perspective, the figure below shows the graphs of and Note the intersection points.

3. Solving Trigonometric Equations With Multiple Angles

So far, we have solved trigonometric equations where the angle is x. How do things change when the angle is a multiple of x, such as or

Consider the function The period of this function is meaning it goes through one full oscillation from to As a result, the graph goes through two full oscillations to which is the standard interval over which we solve trigonometric equations.

Note that when the angle is Thus, from to the angle of the expression starts at and ends at which also represents two full rotations on the unit circle.

This is the logic we use to solve trigonometric equations that contain multiple angles that cannot be manipulated through identities.

EXAMPLE

Consider the equation where

The given domain is

Multiplying each side of the inequality by 2 so that the middle is the domain for angle is

Consider the first rotation of the unit circle. We know that when and when

To get the values from the second rotation, add to each angle from the first rotation since they are coterminal:

and

Thus, we have four values of

Now, divide each angle by 2 to find the values of x  :

Note that each of these solutions is between 0 and

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In this video, we will solve where

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Consider the equation

Find all exact solutions where 0 ≤ θ < 2π.

Note that the angle is Since the given domain is this means that indicating that we first look for solutions from the first two revolutions.

The angles and have a tangent value of 1.

From the first revolution,

From the second revolution, which simplify to

Solving each equation for we have

SOURCE: THIS WORK IS ADAPTED FROM PRECALCULUS BY JAY ABRAMSON. ACCESS FOR FREE AT OPENSTAX.ORG/BOOKS/PRECALCULUS/PAGES/1-INTRODUCTION-TO-FUNCTIONS