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# 2tanx sinx = square root pf 3tanx . find intervals frm o to 2pi?

plzzzzzz helppppp! even the other question i asked b4 this 1!!!

Update:

its from 0 to 2 pie! ( sorry fr writing o)

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2 tanx sinx = √3 tanx

=> 2 tanx sinx - √3 tanx = 0

=> tanx (2sinx - √3) = 0

=> tanx = 0 or sinx = √3/2

=> x = kπ or x = kπ + (-1)^k (π/3)

=> x = { 0, π/3, 2π/3, π, 2π }

taking x ∈ [0, 2π]

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• find intervals that what?

2tan(x)sin(x) = sqrt(3tan(x)). square both sides

4tan^2(x)sin^2(x) = 3tan(x) divide both sides by tan(x)

4tan(x)sin^2(x) = 3 divide both sides by 4

tan(x)sin^2(x) = 3/4

tan is negative in the 2nd and 4th quadrants while sin^2(x) is always non-negative, so we only need consider the first and third quadrants for solution.

in the first quadrant, both sin and tan are increasing, sin(pi/4) = 1/sqrt(2) and tan(pi/4) = 1, so tan(pi/4)sin^2(pi/4) = 1/2

sin(pi/3)= sqrt(3)/2 and tan(pi/2)= sqrt(3) so that tan(pi/3)sin^2(pi/3) = 3/4sqrt(3)

therefore x is in the interval (pi/4, pi/3).

By symmetry, in the third quadrant, all the values are the sam (tan and sin^2 retain their signs)

so that x might also be in the interval (-pi/3, -pi/4)

so

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• hey ishita

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