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TRIGONOMETRIC

EQUATION

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There is no general method in solvingtrigonometric equations.

The following examples shows the solution ofdifferent types of trigonometric equations.

1) Simple trigonometric equations.2) Quadratic equations.

3) Equations involving compound and double-angle-formulae.

4) Equations involving factor formulae.

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5) Equations in the form of a sin b cos = c where a,b,care constants.these equations can be solved by using thefollowing transformations:

a sin + b cos = r sin ( + )

a sin b cos = r sin ( ) a cos + b sin = r cos ( ) a cos + b sin = r cos ( + )

where a > 0 , b > 0

r = a + btan = b a0< < 90

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A.SIMPLE TRIGONOMETRIC EQUATION

1) Solve the following equations for 0 < < 360

i. cot = 3

tan = 1/ 3

= 30, ( 180 + 30 ) = 30, 210

ii. cot 2 = 1/2

2 = 45, ( 360- 45) , ( 360 + 45 ) , (720 45 )2 = 45, 315, 405, 675

= 22 , 157 , 202 , 337

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2) Solve the following sinx = -0.4

* Since sinx is negative,x lies in the third / fourth quadrant.

Case 1: assume that x lies in the third quadrantx = 180 +

sinx = sin ( 180 + )

= -sin

-sin = -0.4

sin = 0.4

= 23.58

x

= 180 +23.5

8= 203.58

Since sin ( 2n +x ) = sinx = -0.4, we get a set ofanswers,namely

2n + 203.5

8 , n Z.

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Case 2 : assume that x lies in the fourth quadrant.

x= 360

sinx= ( 360 )

= - sin -sin = -0.4

sin = 0.4

= 23.58

x = 360 23.58

= 336.42

Since sin ( 2n +x) = sinx= -0.4, we get a set of answers namely2n + 336.42, n Z.

From cases 1 and 2,we see that the answers to the equation sinx= -0.4 are

x= 2n + 203.58 , 2n + 336.42 , n Z. Notice that in either case 1 or case 2, we are

actually solving sin = 0.4 where is an acute angle. Then letx= 180 + 0 or

x= 360 , ex: + or 2 0. Then,we write the general solutions as :x= 2n + ( + ) or x= 2n + ( 2 )

= (2n +1) + , n Z. = 2 (n + 1) , n Z.

Usually,in questions,xwill be defined. For example, you maybe asked to solve the

equation sinx= -0.4 forxlying in the interval [ 0,2 ]. In this case , the answers are

x= 203.58 andx= 336.42

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ii. 3tan 5 sec + 1 = 0

3(sec 1) 5 sec + 1 = 03 sec 5 sec 2 = 0

3(sec 1) (sec 2) = 0

sec = -1/3 @sec = 2

cos = -3 no solutioncos =

= 60 , 300

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2) Solve the equation 2cos t + sin t = 1, 0 t 2

2 cos t + sin t = 12( 1 - sin t) + sin t = 12 sin t sin t 1 = 0(2 sin t + 1) (sin t 1) = 0

sin t = -1/2 @ sin t = 1

If sin t = 1, then t = 90. Note that we only need to

find t lying in the interval [0, 2]. If sin t = -1/2,then lies either in third or fourth quadrant

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Case 1: Assume t lies in third quadrant

t = 180 +

sin t = sin (180 + sin )= -sin sin =

= 30t = 210

Case 2: Assume t lies in the fourth quadrant

t = 360 sin t = sin ( 360 )

= -sin

sin = = 30t = 330

The answers to the equation 2 cos t + sin t = 1 are 90, 210,330