Number of solutions of the equations sin 2θ + 2 = 4 sinθ + cosθ lying in the interval [π, 5π], is
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Therefore the number of solutions of the given equation is four lying in the interval [π, 5π]
Given : trigonometric equation is sin 2θ + 2 = 4 sinθ + cosθ
To find : The number of solutions of the equation sin 2θ + 2 = 4 sinθ + cosθ lying in the interval [π, 5π].
solution : sin 2θ + 2 = 4 sinθ + cosθ
⇒2sinθ . cosθ + 2 = 4sinθ + cosθ
⇒2sinθ cosθ + 2 - 4sinθ - cosθ = 0
⇒2sinθ (cosθ - 2) - 1(cosθ - 2) = 0
⇒(2sinθ - 1)(cosθ - 2) = 0
⇒(2sinθ - 1) = 0 or, (cosθ - 2) = 0
but cosθ can't be greater than 1.
so, cosθ - 2 ≠ 0 ⇒cosθ ≠ 2
hence, 2sinθ - 1 = 0 ⇒sinθ = 1/2 = sin(π/6)
θ = nπ + (-1)ⁿ(π/6) , where n is an integer
if n = 1, θ = π - π/6 = 5π/6 < π [ so this is not lying in the interval π to 5π]
if n = 2, θ = 2π + π/6 = 13π/6
if n = 3, θ = 3π - π/6 = 17π/6
if n = 4, θ = 4π + π/6 = 25π/6
if n = 5, θ = 5π - π/6 = 29π/6
therefore are four values of θ lying in the interval [π, 5π] i.e., 13π/6, 17π/6, 25π/6, 29π/6
Therefore the number of solutions of the given equation is four lying in the interval [π, 5π]