# Thread: 3tanx + 2cosx = 0 Solving for x but confused

1. ## 3tanx + 2cosx = 0 Solving for x but confused

Okay, so, I have been given this question to solve for x and get the possible graph solutions. But I looked at part of the answer and don't understand how some of it works.

So the answer is :

3(sinx / cosx) + 2cosx = 0
3sinx + 2cosx / cosx = 0

But then the next line just goes to

3sinx + 2cosx = 0

How can you divide it by cosx and not change the top? I'm so confused!
And then I don't know where I would go from here?

2. ## Re: 3tanx + 2cosx = 0 Solving for x but confused

Consider the following analogy
x /2 = 0
x = 0

Note that the domain of tan(x) doesn't include point where cos(x)=0, so you're multiplying both sides by a non-zero number when you multiply both sides by cos(x)

Misread. Should be this before you conclude the numerator of LHS is 0.
3sin(x) + 2cos^2(x) / cos(x) = 0

3. ## Re: 3tanx + 2cosx = 0 Solving for x but confused Originally Posted by smurfledurple Okay, so, I have been given this question to solve for x and get the possible graph solutions. But I looked at part of the answer and don't understand how some of it works.

So the answer is :

3(sinx / cosx) + 2cosx = 0
3sinx + 2cosx / cosx = 0

But then the next line just goes to

3sinx + 2cosx = 0
What you wrote doesn't become that! But (3sinx+ 2cosx)/cosx= 0 does! Those parentheses are important!

How can you divide it by cosx and not change the top? I'm so confused!
And then I don't know where I would go from here?

4. ## Re: 3tanx + 2cosx = 0 Solving for x but confused

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

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

Let $u=\sin x$

$2u^2-3u-2=0$

$(2u+1)(u-2)=0$

Since there are no solutions to $\sin x=2$ it must be that $\sin x =-\dfrac 1 2$.

So $x=\dfrac{7\pi}{6}+2n\pi$ or $x=\dfrac{11\pi}{6}+2n\pi$.

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