Higher Engineering Mathematics, Sixth Edition

158 Higher Engineering Mathematics

i.e. 18+18tan^2 A−3tanA− 21 = 0

18tan^2 A−3tanA− 3 = 0

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

Hence 6tanA− 3 =0, from which, tanA=^36 =0.5000

or 3tanA+ 1 =0, from which, tanA=−^13 =− 0 .3333.

Thus A=tan−^1 ( 0. 5000 )= 26. 57 ◦or 206. 57 ◦,since

tangent is positive in the first and third quadrants, or

A=tan−^1 (− 0. 3333 )= 161. 57 ◦or 341. 57 ◦, since tan-

gent is negative in the second and fourth quadrants.

Hence,

A= 26. 57 ◦, 161. 57 ◦, 206. 57 ◦or 341. 57 ◦

Problem 16. Solve 3cosec^2 θ− 5 =4cotθin the

range 0<θ < 360 ◦.

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

3cosec^2 θ− 5 =4cotθgives:

3 (cot^2 θ+ 1 )− 5 =4cotθ

3cot^2 θ+ 3 − 5 =4cotθ

3cot^2 θ−4cotθ− 2 = 0

Since the left-hand side does not factorize 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. 17 ◦ or 210. 17 ◦, since cotangent

is positive in the first and third quadrants, or

θ=cot−^1 (− 0. 3874 )= 111. 18 ◦ or 291. 18 ◦,since

cotangent is negative in the secondand fourthquadrants.

Hence,

θ= 30. 17 ◦, 111. 18 ◦, 210. 17 ◦or 291. 18 ◦

Now try the following exercise

Exercise 69 Further problems on

trigonometric equations

In Problems 1 to 12 solve the equations for angles

between 0◦and 360◦.

1. 2cos^2 θ+sinθ= 1
   [θ= 90 ◦, 210 ◦, 330 ◦]


2. 4cos^2 t+5sint= 3
   [t= 190. 1 ◦, 349. 9 ◦]


3. 2cosθ−4sin^2 θ= 0
   [θ= 38. 67 ◦, 321. 33 ◦]


4. 3cosθ+2sin^2 θ= 3
   [θ= 0 ◦, 60 ◦, 300 ◦, 360 ◦]


5. 12sin^2 θ− 6 =cosθ[
   θ= 48. 19 ◦, 138. 59 ◦,
   221. 41 ◦or 311. 81 ◦

]

6. 16secx− 2 =14tan^2 x
   [x= 52. 53 ◦or 307. 07 ◦]


7. 4cot^2 A−6cosecA+ 6 =0[A= 90 ◦]


8. 5sect+2tan^2 t= 3
   [t= 107. 83 ◦or 252. 17 ◦]


9. 2.9cos^2 a−7sina+ 1 = 0
   [a= 27. 83 ◦or 152. 17 ◦]


10. 3cosec^2 β= 8 −7cot[β
    β= 60. 17 ◦, 161. 02 ◦,


11. 17 ◦or 341. 02 ◦

]

11. cotθ=sinθ [51. 83 ◦, 308. 17 ◦]


12. tanθ+3cotθ=5secθ [30◦, 150 ◦]