Higher Engineering Mathematics

TRIGONOMETRIC IDENTITIES AND EQUATIONS 169

B

Problem 7. Solve 1 .5 tanx− 1. 8 = 0 for

0 ◦≤x≤ 360 ◦.

1 .5 tanx− 1. 8 =0, from which

tanx=

1. 8

1. 5

= 1 .2000.

Hencex=tan−^1 1.2000.
Tangent is positive in the first and third quadrants
(see Fig. 16.4) The acute angle tan−^11. 2000 =
50 ◦ 12 ′. Hence,

x= 50 ◦ 12 ′or 180◦+ 50 ◦ 12 ′= 230 ◦ 12 ′

Figure 16.4

Problem 8. Solve 4 sect=5 for values oft

between 0◦and 360◦.

4 sect=5, from which sect=^54 = 1 .2500.

Hencet=sec−^1 1.2500.
Secant=(1/cosine) is positive in the first and
fourth quadrants (see Fig. 16.5) The acute angle
sec−^11. 2500 = 36 ◦ 52 ′. Hence,

t= 36 ◦ 52 ′or 360◦− 36 ◦ 52 ′= 323 ◦ 8 ′

Figure 16.5

Now try the following exercise.

Exercise 74 Further problems on trigono-

metric equations

In Problems 1 to 3 solve the equations for angles

between 0◦and 360◦.

1. 4−7 sinθ=0[θ= 34 ◦ 51 ′or 145◦ 9 ′]


2. 3 cosecA+ 5. 5 = 0
   [A= 213 ◦ 3 ′or 326◦ 57 ′]


3. 4(2. 32 − 5 .4 cott)= 0
   [t= 66 ◦ 45 ′or 246◦ 45 ′]

16.5 Worked problems (ii) on

trigonometric equations

Problem 9. Solve 2−4 cos^2 A=0 for values

ofAin the range 0◦<A< 360 ◦.

2 −4 cos^2 A=0, from which cos^2 A=^24 = 0. 5000

Hence cosA=

√

(0.5000)=± 0 .7071 and

A=cos−^1 (± 0 .7071).

Cosine is positive in quadrants one and four and

negative in quadrants two and three. Thus in this

case there are four solutions, one in each quadrant

(see Fig. 16.6). The acute angle cos−^10. 7071 = 45 ◦.

Hence,

A= 45 ◦, 135 ◦, 225 ◦or 315◦

Problem 10. Solve 21 cot^2 y= 1 .3 for

0 ◦<y< 360 ◦.