Higher Engineering Mathematics, Sixth Edition

Trigonometric identities and equations 157

15.6 Worked problems (iii)on

trigonometric equations

Problem 12. Solve the equation

8sin^2 θ+2sinθ− 1 = 0 ,

for all values ofθbetween 0◦and 360◦.

Factorizing 8sin^2 θ+2sinθ− 1 =0gives
(4sinθ− 1 )(2sinθ+ 1 )=0.
Hence 4sinθ− 1 =0, from which, sinθ=^14 = 0 .2500,

or 2sinθ+ 1 =0, from which, sinθ=−^12 =− 0 .5000.
(Instead of factorizing, the quadratic formula can, of
course, be used).
θ=sin−^10. 2500 = 14. 48 ◦ or 165. 52 ◦,sincesine
is positive in the first and second quadrants, or
θ=sin−^1 (− 0. 5000 )= 210 ◦or 330◦, since sine is neg-
ative in the third and fourth quadrants. Hence

θ= 14. 48 ◦, 165. 52 ◦, 210 ◦or 330◦

Problem 13. Solve 6cos^2 θ+5cosθ− 6 =0for

values ofθfrom 0◦to 360◦.

Factorizing 6cos^2 θ+5cosθ− 6 =0gives
(3cosθ− 2 )(2cosθ+ 3 )=0.
Hence 3cosθ− 2 =0, from which, cosθ=^23 =0.6667,

or 2cosθ+ 3 =0, from which, cosθ=−^32 =− 1 .5000.

The minimum value of a cosine is−1, hence the lat-
ter expression has no solution and is thus neglected.
Hence,

θ=cos−^10. 6667 = 48. 18 ◦or 311. 82 ◦

since cosine is positive in the first and fourth quadrants.

Now try the following exercise

Exercise 68 Further problems on

trigonometric equations

In Problems 1 to 3 solve the equations for angles

between 0◦and 360◦.

1. 15sin^2 A+sinA− 2 =[ 0
   A= 19. 47 ◦, 160. 53 ◦,


2. 58 ◦or 336. 42 ◦

]

2. 
   

   8tan^2 θ+2tanθ= (^15) [
   θ= 51. 34 ◦, 123. 69 ◦,

   
   


3. 
   

   34 ◦or 303. 69 ◦
   ]

   
   


4. 
   

   2cosec^2 t−5cosect=[ 12
   t= 14. 48 ◦, 165. 52 ◦,
   221. 81 ◦or 318. 19 ◦

   
   

]

4. 2cos^2 θ+9cosθ− 5 = 0
   [θ= 60 ◦or 300◦]

15.7 Worked problems (iv) on

trigonometric equations

Problem 14. Solve 5cos^2 t+3sint− 3 =0for

values oftfrom 0◦to 360◦.

Sincecos^2 t+sin^2 t= 1 ,cos^2 t= 1 −sin^2 t.Substituting

for cos^2 tin 5cos^2 t+3sint− 3 =0gives:

5 ( 1 −sin^2 t)+3sint− 3 = 0

5 −5sin^2 t+3sint− 3 = 0

−5sin^2 t+3sint+ 2 = 0

5sin^2 t−3sint− 2 = 0

Factorizing gives (5sint+ 2 )(sint− 1 )=0. Hence

5sint+ 2 =0, from which, sint=−^25 =− 0 .4000, or

sint− 1 =0, from which, sint=1.

t=sin−^1 (− 0. 4000 )= 203. 58 ◦or 336. 42 ◦,sincesine

is negative in the third and fourth quadrants, or

t=sin−^11 = 90 ◦. Hencet= 90 ◦, 203. 58 ◦or 336. 42 ◦

as shown in Fig. 15.7.

1.0

y

t 8

2 0.4

2 1.0

0908

203.58 8

y 5 sin t

336.42 8

2708 3608

Figure 15.7

Problem 15. Solve 18sec^2 A−3tanA=21 for

values ofAbetween 0◦and 360◦.

1 +tan^2 A=sec^2 A. Substituting for sec^2 A in

18sec^2 A−3tanA=21 gives

18 ( 1 +tan^2 A)−3tanA=21,