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maths- trigonometry

how do you solve : 2cosx = 3tanx in the interval 0< x < 2pi
Reply 1
Rewrite tan(x)\tan(x) as sin(x)cos(x)\dfrac{\sin(x)}{\cos(x)}

Then multiply the equation by cosx and make use of sin2(x)+cos2(x)=1\sin^{2}(x)+\cos^{2}(x)=1 to get a quadratic to be solved in sinx.
Reply 2
hi, hope this helps as a solution:

tanx=sinx/cosx
substitute this in 2cosx=3sinx/cosx
times both sides by cosx

2cos(^2)x=3sinx

cos(^2)x=1-sin(^2)x

sub this in: 2-2sin(^2)x=3sinx

rearrange so 2sin(^2)x+3sinx-2=0

solve as a quadratic 2x^2+3x-2=0
(x+2)(2x-1)=0
sinx=-2 or sinx=0.5
^no solution

x=sin-1(0.5)
x=(1/6)pi or x=(5/6)pi

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