6.04 Trigonometric equations

The method for solving trigonometric equations is roughly the same. After any algebraic manipulation use the positive value on the right-hand side of the equation to find the size of the related acute angle. Then, draw a quadrant (ASTC) diagram to locate the angles you require to make the equation true. Write down your solutions, ensuring that they are in the domain specified in the question. The sign (positive or negative) and the trig function will determine which quadrants the solutions are in.

There are many variations on trigonometric equations you need to be comfortable with. In this set, we will consider five types of equations.

Type 1: equations involving exact values

• If a question doesn't specify a need for your final answer to be rounded, this indicates that it involves an exact value solution.

Type 2: equations involving boundary angles

• Questions, where the value specified, is 0 or \pm 1 land on what we can refer to as boundary angles. It might be easier for you to solve these questions by considering the sketch of the trigonometric function in question, rather than trying to use your calculator or a quadrant diagram. The exception to this is when \tan\, x equals to \pm 1, where it's more valuable to consider the exact value triangles to find a solution.

Type 3: equations with a different domain

• It doesn't matter how you arrive at your final solutions here, as long as you get there in the end. Many students prefer to solve an equation as if the domain was 0\degree \leq x \leq 360\degree and then "move" these answers into the correct domain by adding or subtracting 360\degree until they find the solutions required. You might feel comfortable looking at your quadrant diagram after finding the relative acute angle and going straight to writing the solutions in the required domain.
• Having a variety of domains might mean you have more or less than two solutions, which is typically what we expect when solving for 0\degree \leq x \leq 360\degree.

Examples

Example 1

Solve \sin x=\dfrac{\sqrt{3}}{2} for x where 0\degree \leq x \leq 90\degree.

Worked Solution

Create a strategy

Use the exact value triangles.

Apply the idea

Using this exact value triangle, we can see that the acute value of x = 60\degree since \\ \sin 60\degree = \dfrac{\sqrt{3}}{2}.

Since the domain is 0\degree \leq x \leq 90\degree we do not need to consider any of the other quadrants, so the only solution is: x=60\degree

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Example 2

Solve \tan x =\dfrac{1}{\sqrt{3}}\, for x where 0\degree \leq x \leq 360\degree .

Worked Solution

Create a strategy

Consider the sign to find the quadrants x could be in. Then use exact value triangles to solve the equation.

Apply the idea

Using this exact value triangle, we can see that the acute value of x = 30\degree since \\ \tan 30\degree = \dfrac{1}{\sqrt{3}}.

We need \tan\,x to be positive since \dfrac{1}{\sqrt{3}} \gt 0. This quadrant diagram shows us that we need to find equivalent angles in the first and third quadrant.

The angle in the first quadrant is 30\degree , the angle in the third quadrant is 180\degree +30\degree=210\degree .

So, the solutions are: x = 30\degree ,\,210\degree

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