Higher Engineering Mathematics

170 GEOMETRY AND TRIGONOMETRY

Figure 16.6

1

2 cot

(^2) y= 1 .3, from which, cot (^2) y=2(1.3)= 2 .6.
Hence coty=
√
2. 6 =± 1 .6125, and y= cot−^1
(± 1 .6125). There are four solutions, one in each
quadrant. The acute angle cot−^11. 6125 = 31 ◦ 48 ′.
Hencey= 31 ◦ 48 ′, 148 ◦ 12 ′, 211 ◦ 48 ′or 328◦ 12 ′.
Now try the following exercise.
Exercise 75 Further problems on trigono-
metric equations
In Problems 1 to 3 solve the equations for angles
between 0◦and 360◦.

1. 5 sin^2 y= 3
   [
   y= 50 ◦ 46 ′, 129◦ 14 ′,
   230 ◦ 46 ′or 309◦ 14 ′

]

2. 5+3 cosec^2 D= 8

[D= 90 ◦or 270◦]

3. 2 cot^2 θ= 5
   [
   θ= 32 ◦ 19 ′, 147◦ 41 ′,

212 ◦ 19 ′or 327◦ 41 ′

]

16.6 Worked problems (iii) on

trigonometric equations

Problem 11. Solve the equation

8 sin^2 θ+2 sinθ− 1 =0,

for all values ofθbetween 0◦and 360◦.

Factorising 8 sin^2 θ+2 sinθ− 1 =0gives

(4 sinθ−1) (2 sinθ+1)=0.

Hence 4 sinθ− 1 =0, from which, sinθ=^14 =

0 .2500, or 2 sinθ+ 1 =0, from which, sinθ=−^12 =

− 0 .5000. (Instead of factorising, the quadratic for-

mula can, of course, be used).

θ=sin−^10. 2500 = 14 ◦ 29 ′or 165◦ 31 ′, since sine

is positive in the first and second quadrants, or

θ=sin−^1 (− 0 .5000)= 210 ◦or 330◦, since sine is

negative in the third and fourth quadrants. Hence

θ= 14 ◦ 29 ′, 165 ◦ 31 ′, 210 ◦or 330◦

Problem 12. Solve 6 cos^2 θ+5 cosθ− 6 = 0

for values ofθfrom 0◦to 360◦.

Factorising 6 cos^2 θ+5 cosθ− 6 =0gives

(3 cosθ−2) (2 cosθ+3)=0.

Hence 3 cosθ− 2 =0, from which, cosθ=^23 =

0.6667, or 2 cosθ+ 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 ◦ 11 ′or 311◦ 49 ′

since cosine is positive in the first and fourth

quadrants.

Now try the following exercise.

Exercise 76 Further problems on trigono-

metric equations

In Problems 1 to 3 solve the equations for angles

between 0◦and 360◦.

1. 15 sin^2 A+sinA− (^2) [= 0
   A= 19 ◦ 28 ′, 160◦ 32 ′,
   203 ◦ 35 ′or 336◦ 25 ′
   ]


2. 8 tan^2 θ+2 tanθ= (^15) [
   θ= 51 ◦ 20 ′, 123◦ 41 ′,
   231 ◦ 20 ′ or 303◦ 41 ′
   ]