Trig functions Q :) (1 Viewer)

NewiJapper · May 4, 2010

Solve for x: $sin^{2}x-3sinxcosx+2cos^{2}x=0$ , where $0\leq x\leq 2pi$

With the worked solution they put everything over $cos^{2}x$ but i dont understand why

Help/answer would be appreciated!

P.s with an explanation if you could because i really dont understand the worked solution :S

mmo · May 4, 2010

hi mate
this is quick reply
you divide everything by cos2 so you can get a cancellation of a TAN and a 1 and then it is a quardatic eqation that you can solve

thanks

js992 · May 4, 2010

Because you have cosx and sin x in the equation, you cannot easily solve it by factorising. Ideally, you only want one trig function, ie, all sin or all cos or all tan.

By dividng through by $cos^{2}x$ , you end up with
$tan^{2}x-3tanx+2=0$

Now you can solve easily by letting tan x = x and solving it as though it was a quadratic.

$x^2-3x+2=0$
$(x-2)(x-1)=0$
$x=2 or x=1$
$tanx =2$ or 1

Now solve for x to get your answer

NewiJapper · May 4, 2010

thanks guys!

Drongoski · May 4, 2010

This is a quadratic eqn in sinx and cos x So can be solved this way:

$sin^{2}x-3sinxcosx+2cos^{2}x\ =\ 0\\ \therefore\ \ (sinx-cosx)(sinx-2cosx)\ =\ 0\\ \\ \therefore \ \ either\ \ sinx-cosx\ =\ 0\ \ \Rightarrow \ \ \frac{sinx}{cosx}=tanx=1\\ \Rightarrow \ \ x=\frac{\pi }{4}\ \ or\ \ \frac{5\pi }{4}\\ \\ or:\ \ sinx-2cosx\ =\ 0\ \ \Rightarrow \ \ \frac{sinx}{cosx}=tanx\ =\ 2\\ \Rightarrow \ \ x\ =\ you\ work\ it\ out.$

Trig functions Q :) (1 Viewer)

NewiJapper

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mmo

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js992

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NewiJapper

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Drongoski

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