If 2sin⁡θ+cos⁡θ=3, what is the value of tan2θ+4tan⁡θ ?

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Melinda McCombs · Accepted Answer

(2sin⁡θ+cos⁡θ)2=3  4sin2θ+cos2θ+4sin⁡θcos⁡θ=3sin2θ+3cos2θ   sin2θ+4sin⁡θcos⁡θ=2cos2θ   Now dividing both sides by cos2θ   tan2θ+4tan⁡θ=2

yotaniwc · Accepted Answer

Alternative: We have2sin⁡θ=3−cos⁡θSquaring,4−4cos2θ=3+cos2θ−23cos⁡θSo,5cos2θ−23cos⁡θ−1=0Orcos⁡θ=23±4210Since further conditions haven&#039;t been given, both values would be valid, and from here tan⁡θ can be calculated.

karton · Accepted Answer

Simply square the constraint and you get $4 \sin^{2} (x) + 4 \sin (x) \cos (x) + \cos^{2} (x) = 3$ then double angle and trig identity $3 \sin^{2} (x) + 2 \sin (2 x) = 2$ furthermore this is equivalent to $2 \sin (2 x) = 3 \cos^{2} (x) - 1$
and with your definition of the left hand side
$\tan^{2} x + 4 \tan x = \frac{\sin^{2} x + 2 \sin 2 x}{\cos^{2} x}$
we can insert the definition of $2 \sin (2 x)$ we got from the constraint and we get
$\tan^{2} x + 4 \tan x = \frac{\sin^{2} x + 3 \cos^{2} x - 1}{\cos^{2} x} = \frac{2 \cos^{2} x}{\cos^{2} x} = 2$

If 2 \sin \theta+\cos \theta=\sqrt3, what is the value of

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