Question

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

Answer 1

`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)`.

Answer 2

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.