Lesson

Trigonometric equations, such as $\tan x=\sqrt{3}$`t``a``n``x`=√3, require us to find the size of an angle $x$`x` that results in the given value. Unlike other equations we have solved, such as linear or quadratic equations, we are faced with a problem. This is because an equation like $\tan x=\sqrt{3}$`t``a``n``x`=√3 has an infinite number of solutions.

If you think back to the graph of $y=\tan x$`y`=`t``a``n``x`, you will recall that there will be two angles in every $360^\circ$360° cycle of the graph that will have a $y$`y`-value of $\sqrt{3}$√3. Remembering our work with angles of any magnitude on the unit circle, you will know that these values occur in the first and third quadrant (the "A" and "T" of ASTC). The diagram below illustrates this:

Graph of $y=\tan x$y=tanx and $y=\sqrt{3}$y=√3 |

For this reason, a trigonometric equation will always be accompanied by another piece of information: the domain within which we are trying to find solutions. Most of the time, this will be the domain $0^\circ\le x\le360^\circ$0°≤`x`≤360°. Within this domain, it is reasonable to expect that we will find two solutions to a typical trigonometric equation.

The method for solving trigonometric equations is roughly the same. After any algebraic manipulation, you might need to undertake, 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.

Solve $\tan\theta=\sqrt{3}$`t``a``n``θ`=√3 in the domain $0^\circ\le\theta\le360^\circ$0°≤`θ`≤360°.

**Think:** Let's first find the relative acute angle that satisfies this equation, and then use this angle to find the remaining solutions in the given domain.

**Do:** Using our knowledge of exact value triangles, we have that:

$\tan\theta$tanθ |
$=$= | $\sqrt{3}$√3 | (Given) |

$\theta$θ |
$=$= | $60^\circ$60° | (Using our knowledge of exact value triangles) |

So $\theta=60^\circ$`θ`=60° is our relative acute angle. We need $\tan\theta$`t``a``n``θ` to be positive, so our quadrant diagram below shows us that we need to find angles in the first and third quadrant:

Hence, the two solutions will be the acute relative angle we found, $60^\circ$60°, and the equivalent relative angle in the third quadrant, which is $180^\circ+60^\circ$180°+60°. So our solutions are:

$\theta$θ |
$=$= | $60^\circ,240^\circ$60°,240° |

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$0 or $\pm1$±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 $\tan x$`t``a``n``x` equals to $\pm1$±1, where it's more valuable to consider the exact value triangles to find a soluiton.

**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^\circ\le x\le360^\circ$0°≤`x`≤360° and then "move" these answers into the correct domain by adding or subtracting $360^\circ$360° 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^\circ\le x\le360^\circ$0°≤`x`≤360°.

Solve $\sin x=\frac{1}{2}$`s``i``n``x`=12 for $x$`x` where $0^\circ\le x\le90^\circ$0°≤`x`≤90°.

Consider the equation $\sin\theta=\frac{\sqrt{3}}{2}$`s``i``n``θ`=√32 for $0^\circ<\theta<180^\circ$0°<`θ`<180°.

How many solutions for $\theta$

`θ`does the equation have?Zero

AOne

BTwo

CInfinitely many

DZero

AOne

BTwo

CInfinitely many

D

Solve $\cos x=\frac{1}{2}$`c``o``s``x`=12 for $x$`x` where $0^\circ\le x\le360^\circ$0°≤`x`≤360°.

Enter all solutions on the same line, separated by commas.

Solve simple trigonometric equations.