TRIGONOMETRIC EQUATIONS TEACHER NOTE: Only degrees are used in these slides. Questions are also asked on radians, so teachers should make sure to show how to solve questions using radians in other examples.

Firstly recall the graphs of y = sin x, y = cos x and y = tan x. Note: sin x is positive for 0º < x < 180º, and negative for 180º < x < 360º. y = sin x Note: cos x is positive for 0º ≤ x < 90º, and for 270º < x ≤ 360º and negative for 90º < x < 270º. y = cos x y = tan x Note: tan x is positive for 0° < x < 90º, and for 180º < x < 270º and negative for 90º < x < 180º. and for 270º < x < 360º.

S A T C We can summarise the information in the following diagram: 0º, 360º 180º 90º 270º sin +ve cos +ve tan +ve sin +ve cos –ve tan –ve sin –ve cos –ve tan +ve sin –ve cos +ve This can be simplified to show just the positive ratios: 0º, 360º 180º 90º 270º all +ve sin +ve tan +ve cos +ve A S T C TEACHER NOTE: CAST diagram used here – teacher could also use sin, cos and tan graphs here. or just:

S A T C The positive ratio diagram can be used to solve trigonometric equations: Example 1: Solve cos x = 0.5; 0º ≤ x < 360º A S T C We can see that the cosine of an angle is positive in the two quadrants on the right, i.e. the 1st and 4th quadrants. Using:  α = cos–1 0.5 = 60º Hence the solutions in the given range are: x = 60°, 360 – 60° so x = 60º, 300º

Example 2: Solve sin x = –0.5; –360º < x ≤ 360º The sine of an angle is negative in the 3rd and 4th quadrants.  α = sin–1 0.5 = 30º N.B. The negative is ignored to find the acute angle. Hence the solutions in the given range are: x = –150°, –30°, 210°, 330° In problems where the angle is not simply x, the given range will need to be adjusted.

Example 3: Solve 3 + 5 tan 2x = 0; 0º ≤ x ≤ 360º. Firstly we need to make tan 2x the subject of the equation: – 3 5 tan 2x = The tangent of an angle is negative in the 2nd and 4th quadrants.  = 30.96° The range must be adjusted for the angle 2x. i.e. 0° ≤ 2x ≤ 720°. Hence: 2x = 149.04°, 329.04°, 509.04°, 689.04°. x = 74.5°, 164.5°, 254.5°, 344.5°.

Example 4: Solve 2 sin (4x + 90º) – 1 = 0; 0 < x < 90º The sine of an angle is positive in the 1st and 2nd quadrants.  = 30º The range must be adjusted for the angle 4x + 90º. i.e. 0º < 4x < 360º 90º < 4x + 90º < 450º 4x + 90º = 150°, 390° 4x = 60º, 300º x = 15º, 75º

Example 5: Solve 6 sin2 x + sin x – 1 = 0; 0º ≤ x < 360º A quadratic equation! It may help to abbreviate sin x with s: i.e. 6s2 + s – 1 = 0 Factorising this: (3s – 1)(2s + 1)= 0   α = 19.47º α = 30º So, x = 19.5°, 160.5°, 210°, 330°. (To nearest 0.1º.)

A S T C Summary of key points: To solve a Trigonometric Equation: Re-arrange the equation to make sin, cos or tan of some angle the subject. Locate the quadrants in which the ratio is positive, or negative as required. A S T C Using: Adjust the range for the given angle. Read off all the solutions within the range.