156 Higher Engineering Mathematics
In Problems 4 to 6, solve forθ in the range
0 ◦≤θ≤ 360 ◦.- secθ=2[ 60 ◦, 300 ◦]
- cotθ= 0 .6[59◦, 239 ◦]
- cosecθ= 1 .5[ 41. 81 ◦, 138. 19 ◦]
In Problems 7 to 9, solve forx in the range
− 180 ◦≤x≤ 180 ◦.- secx=− 1 .5[± 131. 81 ◦]
- cotx= 1 .2[39. 81 ◦,− 140. 19 ◦]
- cosecx=−2[− 30 ◦,− 150 ◦]
In Problem 10 and 11, solve forθ in the range
0 ◦≤θ≤ 360 ◦.- 3sinθ=2cosθ [33. 69 ◦, 213. 69 ◦]
- 5cosθ=−sinθ [101. 31 ◦, 281. 31 ◦]
15.5 Worked problems (ii)on
trigonometric equations
Problem 10. Solve 2−4cos^2 A=0 for values of
Ain the range 0◦<A< 360 ◦.2 −4cos^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 neg-
ative in quadrants two and three. Thus in this case there
are four solutions, one in each quadrant (see Fig. 15.6).
The acute angle cos−^10. 7071 = 45 ◦. Hence,A= 45 ◦, 135 ◦, 225 ◦or 315◦Problem 11. Solve^12 cot^2 y= 1 .3for
0 ◦<y< 360 ◦.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. 81 ◦.
Hencey= 31. 81 ◦, 148. 19 ◦, 211. 81 ◦or 328. 19 ◦.
1.0
y
A 8
y 5 cos A
0
0.7071
1358
458 1808 31583608
2258
2 0.7071
2 1.0
(a)
(b)
458
458 458
TC
S A
458
1808
3608
2708
908
0
Figure 15.6
Now try the following exercise
Exercise 67 Further problems on
trigonometric equations
In Problems 1 to 3 solve the equations for angles
between 0◦and 360◦.
- 5sin^2 y= 3
[
y= 50. 77 ◦, 129. 23 ◦, - 77 ◦or 309. 23 ◦
]- cos^2 θ= 0. 25
[θ= 60 ◦, 120 ◦, 240 ◦or 300◦] - tan^2 x= 3
[θ= 60 ◦, 120 ◦, 240 ◦or 300◦] - 5+3cosec^2 D= 8
[D= 90 ◦or 270◦] - 2cot^2 θ= 5
[
θ= 32. 32 ◦, 147. 68 ◦,
212. 32 ◦or 327. 68 ◦
]