Higher Engineering Mathematics, Sixth Edition

154 Higher Engineering Mathematics

3. 2cos^2 A− 1 =cos^2 A−sin^2 A

4.

cosx−cos^3 x

sinx

=sinxcosx

5. ( 1 +cotθ)^2 +( 1 −cotθ)^2 =2cosec^2 θ

6.

sin^2 x(secx+cosecx)

cosxtanx

= 1 +tanx

15.3 Trigonometric equations

Equations which contain trigonometric ratios are called

trigonometric equations. There are usually an infinite

number of solutions to such equations; however, solu-

tions are often restricted to those between 0◦and 360◦.

A knowledge of angles of any magnitude is essential

in the solution of trigonometric equations and calcula-

tors cannot be relied upon to give all the solutions (as

shown in Chapter 14). Fig. 15.2 shows a summary for

angles of any magnitude.

908

1808

2708

3608

08

Sine

(and cosecant

positive)

Tangent

(and cotangent

positive)

Cosine

(and secant

positive)

All positive

Figure 15.2

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 positive.

Since when a square root is taken there is a pos-

itive and negative answer there are four values

ofAbetween 0◦and 360◦which satisfy 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 equation

in which the unknown is sinA. The solution of

a quadratic equation is obtained either by fac-

torizing (if possible) or by using the quadratic

formula:

sin A=

−b±

√

(b^2 −4ac)

2a

(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).

15.4 Worked problems (i) on

trigonometric equations

Problem 6. Solve the trigonometric equation

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

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

Henceθ=sin−^1 (− 0. 6000 ). Sineis negative in thethird

and fourth quadrants (see Fig. 15.3). The acute angle

sin−^1 ( 0. 6000 )= 36. 87 ◦(shown asαin Fig. 15.3(b)).

Hence,

θ= 180 ◦+ 36. 87 ◦,i.e. 216. 87 ◦ or

θ= 360 ◦− 36. 87 ◦,i.e. 323. 13 ◦

Problem 7. Solve 1.5tanx− 1. 8 =0for

0 ◦≤x≤ 360 ◦.

1 .5tanx− 1. 8 =0, from which

tanx=

1. 8

1. 5

= 1 .2000.

Hencex=tan−^1 1.2000