Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

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)

3 Answers

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  • 1 decade ago
    Best Answer

    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]

  • holdm
    Lv 7
    1 decade ago

    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

  • 1 decade ago

    hey ishita

    was my previous answer right.

    I will answer on one condition that if my previous answer was right, choose it as the best answer and then I will solve this one for you.

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