Everything Maths Grade 11

17.4 CHAPTER 17. TRIGONOMETRY

Step 1 : Determine in which quadrants the solution lies

We look at the sign ofthe trigonometric function. sin θ is given

as a positive amount ( 0 , 3 ). Reference to the CASTdiagram shows

that sine is positive in the first and second quadrants.

S A

T C

Step 2 : Determine the reference angle

The small angle θ is the angle returned bythe calculator:

sin θ = 0, 3

∴ θ = 17, 46 ◦

Step 3 : Determine the generalsolution

Our solution lies in quadrants I and II. We therefore use θ and

180 ◦− θ, and add the (360◦. n) for the periodicity of sine.

180 ◦−θ θ

180 ◦+θ 360 ◦−θ

I : θ = 17, 46 ◦+ (360◦. n); n∈Z

II : θ = 180◦− 17 , 46 ◦+ (360◦. n); n∈Z

= 162, 54 ◦+ (360◦. n); n∈Z

This is called the general solution.

Step 4 : Find the specific solutions

We can then find all the values of θ by substituting n =.. .− 1; 0; 1; 2;.. .etc.

For example,

If n = 0, θ = 17, 46 ◦;162, 54 ◦

If n = 1, θ = 377, 46 ◦;522, 54 ◦

If n =− 1 , θ =− 342 , 54 ◦;− 197 , 46 ◦

We can find as many aswe like or find specificsolutions in a given interval by

choosing more values for n.

General Solution UsingPeriodicity EMBDI

Up until now we have only solved trigonometricequations where the argument (the bit after the func-
tion, e.g. the θ in cos θ or the (2x−7) in tan(2x−7)), has been θ. If there is anything more complicated
than this we need to bea little more careful.

Let us try to solve tan(2x− 10 ◦) = 2, 5 in the range− 360 ◦≤ x≤ 360 ◦. We want solutions for
positive tangent so using our CAST diagram weknow to look in the 1 stand 3 rdquadrants. Our cal-
culator tells us that 2 x− 10 ◦= 68, 2 ◦. This is our reference angle. So to find the general solution we
proceed as follows:

tan(2x− 10 ◦) = 2, 5

∴ 2 x− 10 ◦ = 68, 2 ◦

I : 2x− 10 ◦ = 68, 2 ◦+ (180◦. n)

2 x = 78, 2 ◦+ (180◦. n)

x = 39, 1 ◦+ (90◦. n); n∈Z