Question

If 6 ΦcosΦ + 2 sin²Φ =5, show that tan²Φ - ⅓

Answer 1

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Answer on Question #86087 - Math - Trigonometry

Question

If $6\Phi \cos \Phi + 2\sin^2 \Phi = 5$, show that $\tan^2 \Phi - \frac{1}{3}$

Solution

$$6 \cos \varphi + 2 \sin^ {2} \varphi = 5;$$ $$6 \cos \varphi + 2 - 2 \cos^ {2} \varphi - 5 = 0;$$ $$6 \cos \varphi - 2 \cos^ {2} \varphi - 3 = 0;$$ $$2 \cos^ {2} \varphi - 6 \cos \varphi + 3 = 0$$

If $x = \cos \varphi$ and $-1 \leq \cos \varphi \leq 1$ that $2x^{2} - 6x + 3 = 0$;

$$2 x ^ {2} - 6 x + 3 = 0;$$ $$x = \frac {6 \pm \sqrt {3 6 - 4 * 2 * 3}}{2 * 2};$$ $$x = 1. 5 - 0. 5 \sqrt {3};$$

$x = 1.5 + 0.5\sqrt{3} - \text{ this solution is not included in the range} - 1 \leq \cos \varphi \leq 1$

Then $\cos \varphi = 1.5 - 0.5\sqrt{3}$

$$\tan^ {2} \varphi + 1 = \frac {1}{\cos^ {2} \varphi};$$

$\tan^2\varphi = \frac{1}{\cos^2\varphi} - 1 = \frac{1}{\left(1.5 - 0.5\sqrt{3}\right)^2} - 1$ and it is not equal to $\frac{1}{3}$.

P.S. The task probably contains a typo $\Phi \cos \Phi$ since that equation can be solved with a help of the graphical method and numerical approximations.