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6cos^2 theta + 5 cos theta -4 = 0
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u can factorise it \[(2\cos \theta -1)(3\cos \theta+4)=0\]
to find the exact solution set?
so u will get \[\theta= \cos^{-1} (1/2) \] and \[\theta = \cos^{-1} (-4/3)\]
but how is this done?
after factorised. note that 2(cos theta )-1=0 and 3(cos theta) +4=0 rearrange them and solve both equations to get theta.
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ahh, i get it. thanks(:
no prob
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