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

2 Answers
Jun 23, 2018

Answer:

# 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

Answer:

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.