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

Answer 1

Elijah Alexander

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

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)#.

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Answer 2

Carter Anderson

Number of solutions: 5 for #(0, pi)#

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.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.