Higher Engineering Mathematics

168 GEOMETRY AND TRIGONOMETRY

solutions (as shown in Chapter 15). Fig. 16.2 shows
a summary for angles of any magnitude.

Figure 16.2

Equations of the typeasin^2 A+bsinA+c= 0

(i) Whena= 0 ,bsinA+c=0, hence

sinA=−

c

b

andA=sin−^1

(

−

c

b

)

There are two values of Abetween 0◦ and

360 ◦which satisfy such an equation, provided

− 1 ≤

c

b

≤1 (see Problems 6 to 8).

(ii)Whenb= 0 ,asin^2 A+c=0, hence

sin^2 A=−

c

a

, sinA=

√(

−

c

a

)

andA=sin−^1

√(

−

c

a

)

If eitheraorcis a negative number, then

the value within the square root sign is posi-

tive. Since when a square root is taken there is

a positive and negative answer there are four

values ofAbetween 0◦and 360◦which sat-

isfy such an equation, provided− 1 ≤

c

a

≤ 1

(see Problems 9 and 10).

(iii)Whena,bandcare all non-zero:
asin^2 A+bsinA+c=0 is a quadratic equa-
tion in which the unknown is sinA. The solution
of a quadratic equation is obtained either by fac-
torising (if possible) or by using the quadratic
formula:

sinA=

−b±

√

(b^2 − 4 ac)

2 a

(see Problems 11 and 12).

(iv) Often the trigonometric identities

cos^2 A+sin^2 A=1, 1+tan^2 A=sec^2 A and

cot^2 A+ 1 =cosec^2 Aneed to be used to reduce

equations to one of the above forms (see

Problems 13 to 15).

16.4 Worked problems (i) on

trigonometric equations

Problem 6. Solve the trigonometric equation

5 sinθ+ 3 =0 for values ofθfrom 0◦to 360◦.

5 sinθ+ 3 =0, from which sinθ=−^35 =− 0. 6000

Henceθ=sin−^1 (− 0 .6000). Sine is negative in the

third and fourth quadrants (see Fig. 16.3). The

acute angle sin−^1 (0.6000)= 36 ◦ 52 ′(shown asαin

Fig. 16.3(b)). Hence,

θ= 180 ◦+ 36 ◦ 52 ′,i.e. 216 ◦ 52 ′ or

θ= 360 ◦− 36 ◦ 52 ′,i.e. 323 ◦ 8 ′

Figure 16.3