One of the solutions of the equation 8 sin^3 x - 7sin x + √3 cos x = 0 lies in which interval
Answers
you didn't mention the options required here to solve the question.
well, I am solving it without options, just check your options and answer it.
given equation is ....
8sin³x - 7sinx + √3cosx = 0
⇒8sin³x - 6sinx - sinx + √3cosx = 0
⇒-2(3sinx - 4sin³x) + 2(√3/2 cosx - 1/2 sinx) = 0
we know, sin3x = 3sinx - 4sin³x
⇒-2(sin3x) + 2[sin(π/3).cosx - cos(π/3).sinx ] = 0
we know, sinA.cosB - cosA.sinB = sin(A - B)
⇒-2sin(3x) + 2sin(π/3 - x) = 0
⇒sin(π/3 - x) = sin3x
from general solution of trigonometric functions,
x = nπ - 11π/12 , n ∈ Z
x = nπ - 5π/12, n ∈ Z
x = nπ - 2π/3 , n ∈ Z
if we put n = 1 for all x,
we get, π/12 , 7π/12 and π/3
Answer:
x = nπ - 11π/12 , n ∈ Z
x = nπ - 5π/12, n ∈ Z
x = nπ - 2π/3 , n ∈ Z
Step-by-step explanation:
8sin³x - 7sinx + √3cosx = 0
⇒8sin³x - 6sinx - sinx + √3cosx = 0
⇒-2(3sinx - 4sin³x) + 2(√3/2 cosx - 1/2 sinx) = 0
we know, sin3x = 3sinx - 4sin³x
⇒-2(sin3x) + 2[sin(π/3).cosx - cos(π/3).sinx ] = 0
we know, sinA.cosB - cosA.sinB = sin(A - B)
⇒-2sin(3x) + 2sin(π/3 - x) = 0
⇒sin(π/3 - x) = sin3x
from general solution of trigonometric functions,
x = nπ - 11π/12 , n ∈ Z
x = nπ - 11π/12 , n ∈ Zx = nπ - 5π/12, n ∈ Z
x = nπ - 11π/12 , n ∈ Zx = nπ - 5π/12, n ∈ Zx = nπ - 2π/3 , n ∈ Z
if we put n = 1 for all x,
we get, π/12 , 7π/12 and π/3