1. Solving Trigonometric Equations by Algebraic Methods

2. What is the solution of sin  = for 0£  £ 360?
From the figure, we can see that the solutions are
 = 30 or 150

3. What is the solution of sin  = for 0£  £ 360?
Using a calculator, I can only get one of the solutions,  = 30.
Since sin (180 30) = sin 30,
sin 150 =
The other solution is  = 150.

4. Let’s see how to solve sin  = for 0£  £ 360 by algebraic methods.

5. ∵ sin  = > 0 ∴ q may lie in quadrant I or quadrant II. A S C T
Step 1
Determine the quadrants in which  may lie.
sin  = ∵
> 0 C A S T O y x
∴ q may lie in quadrant I
sin  > 0
sin  > 0
or quadrant II.

6. ∵ sin  = > 0 ∴ q may lie in quadrant I or quadrant II. ∵ sin  =
Step 2
Find an acute angle α
such that sin α = .
sin  = ∵
sin  = sin 30

7. sin  = ∵ > 0 ∴ q may lie in quadrant I or quadrant II. sin  = ∵
Step 3
Find the solutions of the equation for each of the quadrants obtained in step 1.
sin  = ∵
sin  = sin 30
Quadrant I II
III IV
Solution α
180 α
180+ α
360 α ∴
 = 30
or 180  30
 ∵  lies in quadrant I or II,
and sin (180  30) = sin 30
 = 30
or 150

8. A S T C Solve cos  = 0.3 for 0£  £ 360. cos  > 0 ∵y x
cos  > 0 ∵
cos  > 0
∴ q may lie in quadrant I
or quadrant IV.
cos  > 0
∵ cos  = 0.3
cos  = cos 72.5 ∴
 = 72.5
or 360  72.5
 ∵  lies in quadrant I or IV, and
cos (360  72.5) = cos 72.5
 = 72.5
or 
(cor. to 1 d.p.)
(cor. to 1 d.p.)

9. Follow-up question 3 - Solve tan  = for . £ 360 0 q C A S Ty x
∵ tan  < 0
tan  < 0
∴ q may lie in quadrant II
or quadrant IV.
tan  < 0
tan  = 3 - ∵
tan  = -tan 60
∴  = 180 - 60
or 360  60
 tan (180 – 60) = – tan 60
tan (360 – 60) = – tan 60
 = 120 or 300

10. In some cases, the trigonometric identities,
and
sin2  + cos2  = 1,
are very useful in solving trigonometric equations.

11. C A S T Solve 4 cos  + sin  = 5 sin   2 cos  for 0£  £ 360.y x
4 cos  + sin  = 5 sin   2 cos 
tan  > 0
4 sin  = 6 cos  2 3
cos
sin =  
cos
sin
tan =
tan  > 0
tan  = 1.5
tan  > 0 ∵
∴ q may lie in quadrant I
or quadrant III.
∴  = 56.3
or 180 
tan (180 ) = tan 56.3
d.p.) 1 to
(cor.
236.3 or
56.3 = 

12. Some trigonometric equations can be reduced to quadratic equations first and then solved.

13. This is a quadratic equation in cos x.
Solve cos2 x + 3 cos x + 2 = 0 for 0£ x £ 360.
This is a quadratic equation in cos x.
cos2 x + 3 cos x + 2 = 0
(cos x + 1)(cos x + 2) = 0
cos x + 1 = or cos x + 2 = 0
cos x = 1 or cos x = 2 (rejected)
x = 180 1
cos - x

14. Follow-up question
Solve for 　.