How to solve this trigonometric equations?tan 2x cot x - 3 = 0 I need to factor this in order to sove for 0 < x < 2pi   thank you

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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tan2x*cotx - 3 = 0

We know that: tan2x = sin2x/cos2x and cotx = cosx/sinx

==> sin2x/cos2x *cosx/sinx = 3

Now we know that sin2x = 2sinx*cosx

==> 2sinxcosx/cos2x * cosx/sinx = 3

Reduce sinx:

==> 2cos^2 x/ cos2x = 3

Now we know that cos2x = 2cos^2 x-1

==> 2cos^2 x/(2cos^2 x -1) = 3

==> 2cos^2 x = 3(2cos^2 x -1)

==> 2cos^2 x = 6cos^2 x - 3

==> -4cos^2 x= -3

==> 4cos^2 x = 3

==> cos^2 x = 3/4

==> cosx = +-sqrt3/ 2

==> x = pi/6, 5pi/6, 7pi/6, and 11pi/6

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