Simple trigonometry equation

Question

The previous class we were doing trigonometry exercises. Before the class finished, our teacher wrote exercises on the table. I am stuck with the following one:

$$ \cos(2x) + 1 + 3\sin x = 0 $$

I have come up with this:

$$ 1= \sin^2 x + \cos^2 x $$ $$ \cos(2x) = \cos^2 x - \sin^2 x $$

When we substitute we get

$$ 2\cos^2 x + 3\sin x = 0 $$

Need to find $x\ldots$

For some basic information about writing math at this site see e.g. here, here, here and here. — wlad, May 27 '15 at 20:01
@BrianTung: Please don't edit unless you're sure you understand the OP's intent; you've turned this from something that was sensible but hard to read into something that's easy to read but makes absolutely no sense. I am fixing this. — WillO, May 27 '15 at 20:02
Use $\cos(2x)=1-2\sin^2 x$. There was a mistake in your calculation. — André Nicolas, May 27 '15 at 20:02
@WillO: Sorry, I did do it too quickly. I re-edited. ETA: To all—sorry, I made a hash of that. — Brian Tung, May 27 '15 at 20:04
trade the $\cos^2 x$ in the last equation you have for $1 - \sin^2 x.$ then use the quadratic formula. — abel, May 27 '15 at 20:07

score 8 · Accepted Answer · edited May 27 '15 at 20:10

8

$$\cos(2x)+1+3\sin x=0$$

now let's substitute $\cos (2x)=1-2\sin^2x$

$2-2\sin^2x+3\sin x=0$

And then, if we denote $y=\sin x$ we get a quadratic equation

$$2-2y^2+3y=0$$

edited May 27 '15 at 20:10

mathlove

139,939

answered May 27 '15 at 20:04

Uri Goren

2,940

This is not the expression that @Triak ask – cosmoscalibur May 27 '15 at 20:11
1

I can solve from this point. Thanks. I couldn't get the idea right. – Triak May 27 '15 at 20:24

Mark Bennet · Answer 2 · 2015-05-27T20:56:37.513

Your final equation was wrong [now corrected], because you have were inaccurate in the substitution. You should have, on your method $$\cos 2x+1+3\sin x=(\cos^2x-\sin^2 x)+(\cos^2x+\sin^2x)+3\sin x$$

Now the terms in $\sin^2 x$ cancel and you get $$2\cos^2 x+3\sin x=0$$

[The original question had $\cos^2 x+3\sin x=0$]

As others have noted, if you have a term in $\sin^2 x$ instead of in $\cos^2 x$ this becomes a quadratic in $\sin x$ which you can solve for $\sin x$ and then for $x$. You do this by using $\cos^2 x=1-\sin^2 x$ if you didn't get there directly.

score 2 · Answer 3 · answered May 27 '15 at 21:38

Although the procedure is almost same as given but I will explain it to the answer $$\cos(2x)+1+3\sin x=0$$ $$2-2\sin^2 x+3\sin x=0$$ $$2\sin^2 x-3\sin x-2=0$$ $$\implies (\sin x-2)(2\sin x+1)=0$$ $$ \implies \sin x-2=0 \implies\sin x=2\quad \text{but}\quad -1\leq\sin x\leq 1 $$ Hence we have $$ 2\sin x+1=0 \implies\sin x=-\frac{1}{2}=\sin\left(\frac{-\pi}{6}\right)$$ Writing the general solution, we get $$ \color {#0b4}{x = n\pi+ \frac{\pi}{6}} \quad \text{&}\quad \color{#0b4} {x=2n\pi- \frac{\pi}{6}} \quad (\forall \quad n\in I) $$ Where, $n$ is an integer.

Simple trigonometry equation

3 Answers3