Question

2 - cos2θ = 3sinθcosθ, sinθ ≠ cosθ then tanθ is??
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Answer 1

Given:

$2-cos2\theta=3sin\theta cos\theta$

To find: $tan\theta$

Step-by-step explanation:

Now, $2-cos2\theta=3sin\theta cos\theta$

$\Rightarrow 2(sin^{2}\theta+cos^{2}\theta)-(cos^{2}\theta-sin^{2}\theta)=3 sin\theta cos\theta$

$\Rightarrow 2sin^{2}\theta+2cos^{2}\theta-cos^{2}\theta+sin^{2}\theta=3 sin\theta cos\theta$

$\Rightarrow 3sin^{2}\theta-3 sin\theta cos\theta+cos^{2}\theta=0$

$\Rightarrow 3 tan^{2}\theta-3 tan\theta+1=0$ since $cos\theta\neq 0$

Using Sridhar Acharya's formula, we get

$\quad tan\theta=\frac{3\pm\sqrt{9-12}}{6}$

$\Rightarrow tan\theta=\frac{3\pm\sqrt{-3}}{6}$

$\Rightarrow tan\theta=\frac{3\pm i\sqrt{3}}{6}$ where $i=\sqrt{-1}$

Answer: $tan\theta=\frac{3\pm i\sqrt{3}}{6}$