Higher Engineering Mathematics

TRIGONOMETRIC IDENTITIES AND EQUATIONS 171

B

3. 2 cosec^2 t−5 cosect=[ 12
   t= 14 ◦ 29 ′, 165◦ 31 ′,
   221 ◦ 49 ′or 318◦ 11 ′

]

16.7 Worked problems (iv) on

trigonometric equations

Problem 13. Solve 5 cos^2 t+3 sint− 3 = 0

for values oftfrom 0◦to 360◦.

Since cos^2 t+sin^2 t=1, cos^2 t= 1 −sin^2 t. Substi-
tuting for cos^2 tin 5 cos^2 t+3 sint− 3 =0 gives:

5(1−sin^2 t)+3 sint− 3 = 0

5 −5 sin^2 t+3 sint− 3 = 0

−5 sin^2 t+3 sint+ 2 = 0

5 sin^2 t−3 sint− 2 = 0

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

5 sint+ 2 =0, from which, sint=−^25 =− 0 .4000,
or sint− 1 =0, from which, sint=1.
t=sin−^1 (− 0 .4000)= 203 ◦ 35 ′ or 336◦ 25 ′, since
sine is negative in the third and fourth quadrants,
or t=sin−^11 = 90 ◦. Hence t= 90 ◦, 203 ◦ 35 ′ or
336 ◦ 25 ′as shown in Fig. 16.7.

Figure 16.7

Problem 14. Solve 18 sec^2 A−3 tanA= 21

for values ofAbetween 0◦and 360◦.

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

18 sec^2 A−3 tanA=21 gives
18(1+tan^2 A)−3 tanA=21,

i.e. 18+18 tan^2 A−3 tanA− 21 = 0

18 tan^2 A−3 tanA− 3 = 0

Factorising gives (6 tanA−3)(3 tanA+1)=0.

Hence 6 tanA− 3 =0, from which, tanA=^36 =

0.5000 or 3 tanA+ 1 =0, from which, tanA=

−^13 =− 0 .3333. ThusA=tan−^1 (0.5000)= 26 ◦ 34 ′

or 206◦ 34 ′, since tangent is positive in the first and

third quadrants, orA=tan−^1 (− 0 .3333)= 161 ◦ 34 ′

or 341◦ 34 ′, since tangent is negative in the second

and fourth quadrants. Hence,

A= 26 ◦ 34 ′, 161 ◦ 34 ′, 206 ◦ 34 ′or 341◦ 34 ′

Problem 15. Solve 3 cosec^2 θ− 5 =4 cotθ

in the range 0<θ< 360 ◦.

cot^2 θ+ 1 =cosec^2 θ. Substituting for cosec^2 θin

3 cosec^2 θ− 5 =4 cotθgives:

3 (cot^2 θ+1)− 5 =4 cotθ

3 cot^2 θ+ 3 − 5 =4 cotθ

3 cot^2 θ−4 cotθ− 2 = 0

Since the left-hand side does not factorise the

quadratic formula is used. Thus,

cotθ=

−(−4)±

√

[(−4)^2 −4(3)(−2)]

2(3)

=

4 ±

√

(16+24)

6

=

4 ±

√

40

6

=

10. 3246

6

or−

2. 3246

6

Hence cotθ= 1 .7208 or −0.3874, θ=cot−^1

1. 7208 = 30 ◦ 10 ′ or 210◦ 10 ′, since cotangent is

positive in the first and third quadrants, or

θ=cot−^1 (− 0 .3874)= 111 ◦ 11 ′ or 291◦ 11 ′, since

cotangent is negative in the second and fourth

quadrants. Hence,

θ= 30 ◦ 10 ′, 111 ◦ 11 ′, 210 ◦ 10 ′or 291◦ 11 ′

Now try the following exercise.

Exercise 77 Further problems on trigono-

metric equations

In Problems 1 to 6 solve the equations for angles

between 0◦and 360◦.

1. 12 sin^2 θ− 6 =cosθ[
   θ= 48 ◦ 11 ′, 138◦ 35 ′,
   221 ◦ 25 ′or 311◦ 49 ′

]