Number of solutions of sinx.tan4x=cosx in (0,π) ?

2 Answers
Jun 23, 2018

x in {pi/10, 3pi/10, pi/2, 7pi/10, 9pi/10} sub (0,pi).

Explanation:

Note that, in the given eqn., sinx!=0, because, otherwise

cosx=sinxtan4x=0," contradicting, "sin^2+cos^2x=1.

Hence, we can divide the eqn. by sinx, and get,

tan4x=cosx/sinx=cotx=tan(pi/2-x).

:. 4x=(pi/2-x)+kpi, i.e., 5x=pi/2+kpi, k in ZZ.

:. x=pi/10+kpi/5=(2k+1)pi/10, k in ZZ.

But, 0 lt x lt pi :. 0 lt (2k+1)pi/10 lt pi

:. 0 lt 2k+1 lt 10 , or, -1/2 lt k lt 9/2, k in ZZ.

:. x=(2k+1)pi/10, k=0,1,2,3,4.

:. x in {pi/10, 3pi/10, pi/2, 7pi/10, 9pi/10} sub (0,pi).

Jun 23, 2018

Number of solutions: 5 for (0, pi)

Explanation:

sin x.tan 4x = cos x
tan 4x = cot x = tan (pi/2 - x)
Unit circle and property of tan give -->
4x = (pi/2 - x) + kpi
5x = pi/2 + kpi
x = pi/10 + (kpi)/5
k = 0 --> x = pi/10
k = 1 --> x = pi/10 + pi/5 = (3pi)/10
k = 2 --> x = pi/10 + (2pi)/5 = (5pi)/10
k = 3 --> x = pi/10 + (3pi)/5 = (7pi)/5
k = 4 --> x = pi/10 + (4pi)/5 = (9pi)/10
In the interval (0, pi), there are 5 answers.