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The question is:

Given that: $$7\sin^2(x) + \sin(x)\cos(x) = 6$$ Show that: $$\tan^2(x) + \tan(x) - 6 = 0$$

I don't know if there's a proof I'm missing when trying to complete this. I've tried both the $\sin(x)/\cos(x) = \tan(x)$ and the $\sin^2(x) + \cos^2(x) = 1$, and neither of them have got me closer to the answer.

Any help would be appreciated because it's driving me insane not knowing how to do this. I'm not sure if I'm missing something in the proofs I've been using or if I need a different one entirely.

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  • $\begingroup$ Using the two identities you know, you can derive another one: tan^2(x)+1=1/cos^2(x). Can you see how to use it here? $\endgroup$
    – Yuriy S
    Jan 23 '21 at 14:28
  • $\begingroup$ Hint: multiply the tan one by $\cos^2$ and then use $\cos^2=1-\sin^2$ (you have to justify why $\cos\neq 0$). $\endgroup$
    – zwim
    Jan 23 '21 at 14:34
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Write as

$$7\sin^2 x + \sin x \cos x = 6(\sin^2 x+\cos^2 x)$$

Simplify to

$$\sin^2 x + \sin x \cos x - 6\cos^2 x=0$$

Show that $\cos x \neq 0$. Hint : $\cos^2 x = 0 \Rightarrow \sin^2 x = 1$.

Divide throughout by $\cos^2 x$,

$$\tan^2 x + \tan x - 6 = 0$$

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    $\begingroup$ As I pointed out in comment, justifying yourself why $\cos(x)\neq 0$ would be nice. $\endgroup$
    – zwim
    Jan 23 '21 at 14:39
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    $\begingroup$ @zwim, Agreed. Edited. $\endgroup$
    – cosmo5
    Jan 23 '21 at 14:47

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