Solving Trigonometric Equations by Algebraic Methods

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

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.

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

∵ 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.

∵ 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

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

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 287.5 (cor. to 1 d.p.) (cor. to 1 d.p.)

Follow-up question 3 - Solve tan  = for . £ 360 0 q C A S T y 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

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

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 + 56.3 tan (180 + 56.3) = tan 56.3 d.p.) 1 to (cor. 236.3 or 56.3 = 

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

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 = 0 or cos x + 2 = 0 cos x = 1 or cos x = 2 (rejected) x = 180 1 cos £ - x

Follow-up question Solve for 　.