Everything Maths Grade 11

17.4 CHAPTER 17. TRIGONOMETRY

SOLUTION

3cos(θ− 15 ◦)− 1 =− 2 , 583

3cos(θ− 15 ◦) =− 1 , 583

cos(θ− 15 ◦) =− 0 , 5276...

reference angle: (θ− 15 ◦) = 58, 2 ◦

II : θ− 15 ◦ = 180◦− 58 , 2 ◦+ (360◦. n); n∈Z

θ = 136, 8 ◦+ (360◦. n); n∈Z

III : θ− 15 ◦ = 180◦+ 58, 2 ◦+ (360◦. n); n∈Z

θ = 253, 2 ◦+ (360◦. n); n∈Z

Quadratic and Higher Order Trigonometric

Equations

EMBDK

The simplest quadratic trigonometric equation isof the form

sin^2 x− 2 =− 1 , 5

This type of equation can be easily solved by rearranging to get a more familiar linear equation

sin^2 x = 0, 5

⇒ sin x =±

�

0 , 5

This gives two linear trigonometric equations. The solutions to either of these equations will satisfythe
original quadratic.

The next level of complexity comes when we need to solve a trinomialwhich contains trigonometric
functions. It is much easier in this case to use temporary variables. Consider solving

tan^2 (2x + 1) + 3tan(2x + 1) + 2 = 0

Here we notice that tan(2x + 1) occurs twice in the equation, hence we let y = tan(2x + 1) and
rewrite:
y^2 + 3y + 2 = 0

That should look rathermore familiar. We canimmediately write downthe factorised form andthe
solutions:

(y + 1)(y + 2) = 0

⇒ y =− 1 OR y =− 2

Next we just substitute back for the temporary variable:

tan(2x + 1) =− 1 or tan(2x + 1) =− 2