Everything Maths Grade 11

17.4 CHAPTER 17. TRIGONOMETRY

infinite number of solutions to this equation! This difficulty (which is caused by the periodicityof the
sine function) makes solving trigonometric equations much harder thanthey may seem to be.

Any problem on trigonometric equations will require two pieces of information to solve. The first is
the equation itself and the second is the range in which your answers must lie. The hard part is making
sure you find all of thepossible answers withinthe range. Your calculator will always give youthe
smallest answer (i.e. the one that lies between− 90 ◦and 90 ◦for tangent and sine andone between 0 ◦
and 180 ◦for cosine). Bearing thisin mind we can alreadysolve trigonometric equations within these
ranges.

Example 8:

QUESTION

Find the values of x for which sin

�x

2

�

= 0, 5 if it is given that x < 90 ◦.

SOLUTION

Because we are told that x is an acute angle, we can simply apply an inverse trigonometric

function to both sides.

sin

�x

2

�

= 0, 5 (17.2)

⇒x 2 = arcsin0, 5 (17.3)

⇒x 2 = 30◦ (17.4)

∴ x = 60◦ (17.5)

We can, of course, solvetrigonometric equationsin any range by drawingthe graph.

Example 9:

QUESTION

For what values of x does sin x = 0, 5 when− 360 ◦≤ x≤ 360 ◦?

SOLUTION

Step 1 : Draw the graph

We take a look at the graph of sin x = 0, 5 on the interval [− 360 ◦;360◦]. We

want to know when the y value of the graph is 0 , 5 so we draw in a line at y = 0, 5.