How To Solve $\sin^2(x)+\sin2x+2\cos^2(x)=0 - Math Stack Exchange

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Learn more about Teams How to solve $\sin^2(x)+\sin2x+2\cos^2(x)=0$ Ask Question Asked 6 years, 4 months ago Modified 6 years, 4 months ago Viewed 657 times 7 $\begingroup$

How do you solve $\sin^2(x)+\sin2x+2\cos^2(x)=0$? I have been able to rewrite it as $(\sin(x)+\cos(x))^2+\cos^2(x)=0$. Not obviously useful, I think

• trigonometry

Share Cite Follow asked Aug 19, 2018 at 1:25 James WarthingtonJames Warthington 3,19711 gold badge1717 silver badges4141 bronze badges $\endgroup$ 3

• 6 $\begingroup$ On the contrary. A sum of two squares happens to be zero in very few cases. Not in yours, since $\sin(x)+\cos(x)$ and $\cos(x)$ cannot vanish at the same time. $\endgroup$ – Jack D'Aurizio Commented Aug 19, 2018 at 1:28
• $\begingroup$ I suppose it can be assumed they want only real solutions? $\endgroup$ – Deepak Commented Aug 19, 2018 at 1:39
• $\begingroup$ Thanks, I got this now. Tons of excellent explanations. $\endgroup$ – James Warthington Commented Aug 19, 2018 at 17:16

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5 Answers 5

Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 13 $\begingroup$

What you've done here is actually perfect! We know that the only way for this expression to be $0$ is if

$$\sin(x)+\cos(x)=0$$ and $$\cos^2(x)=0\to\cos(x)=0$$

We know both can't be possible because that would imply $\sin(x)=\cos(x)=0$, which is false. So the solution of given equation does not exist.

Share Cite Follow edited Aug 19, 2018 at 2:19 Math Lover 15.3k33 gold badges2121 silver badges3838 bronze badges answered Aug 19, 2018 at 1:31 Rushabh MehtaRushabh Mehta 13.8k88 gold badges3030 silver badges4949 bronze badges $\endgroup$ 0 Add a comment | 6 $\begingroup$

Rushabh's answer is perfect, though I would do it slightly differently. Note that $\sin 2x = 2\sin x\cos x$, so your equation can be rewritten as $$ \sin^2 x + 2\sin x\cos x + 2\cos^2 x = 0.$$ If you treat $\sin x$ and $\cos x$ as independent variables $u$ and $v$, this becomes the quadratic equation $$ u^2 + 2uv + 2v^2 = 0, $$ where it becomes clear there are no solutions, as the discriminant is $\triangle = 2^2-4\times 2 = -4<0$.

Share Cite Follow answered Aug 19, 2018 at 2:15 YiFan TeyYiFan Tey 17.7k44 gold badges2929 silver badges6969 bronze badges $\endgroup$ Add a comment | 3 $\begingroup$

From $$(\sin x + \cos x )^2 + \cos ^2x =0$$ we get $$(\sin x + \cos x )^2=0$$ and $$\cos ^2x =0$$

The first identity implies $$\tan x =-1$$ and the second identity implies $$\cos x=0$$ which makes $\tan x = \pm \infty $.

Thus there is no solution which satisfies both identities, that is your equation has no solution.

Share Cite Follow answered Aug 19, 2018 at 2:02 Mohammad Riazi-KermaniMohammad Riazi-Kermani 69.2k44 gold badges4343 silver badges8989 bronze badges $\endgroup$ Add a comment | 2 $\begingroup$

$\sin^2(x) + \cos^2(x) = 1$, so we have:

$$\sin 2x + \cos^2 x + 1 = 0$$

Now, $\cos^2 x ≥ 0$, and $\sin 2x ≥ -1$, so the only solutions are when $\cos^2 x = 0$ and $\sin 2x = -1$.

When $\cos^2 x = 0$, $x = -\frac{\pi}{2} + 2\pi n$ or $x = \frac{\pi}{2} + 2\pi n$. However, $\sin \left( 2(-\frac{\pi}{2}) \right) = 0$ and $\sin \left( 2(\frac{\pi}{2}) \right) = 0$, so there are no solutions.

Share Cite Follow answered Aug 19, 2018 at 1:41 Toby MakToby Mak 17k44 gold badges3131 silver badges4646 bronze badges $\endgroup$ Add a comment | 2 $\begingroup$

Hint:

For $a\sin^2x+b\cos^2x+c\sin2x=0,$

divide both sides by $\cos^2x$ as $\cos x\ne0$(why?)

to find $$at^2+2ct+b=0$$ where $t=\tan x$

For real solution, the discriminant must be $\ge0$

Share Cite Follow answered Aug 19, 2018 at 1:59 lab bhattacharjeelab bhattacharjee 278k2020 gold badges208208 silver badges331331 bronze badges $\endgroup$ Add a comment |

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