Trigonometric Calculus

**metaname90** · November 27th 2010, 07:35 PM

Okay, I always struggle with how to exactly work out the algebra for these trigonometric problems... The question is to identify where and what type of critical points lie on the function:

$f(x)=cos(2x)-1sin(x)$

between 0 and 2pi. Apparently there are 4.

So far this is what I have done:
$f'(x)=-2sin(2x)-cosx$

$0=-4sinxcosx-cosx$

$cosx=-4sinxcosx$

$-4sinx=1$

$sinx=-1/4$

First of all I don't know how to solve for x in the final line without a calculator. Also, I only see that that would only possibly lead to two critical points between 0 and 2pi....

Thank you anyone who can help me a bit!

P.S. I kind of suck at this math formatting so forgive me...

**Mondreus** · November 27th 2010, 07:53 PM

An equation on the form $\cos{v} = c$ has the solutions $v = v_0 + 2\pi n$ and $v = -v_0 + 2\pi n$

An equation on the form $\sin{v} = c$ has the solutions $v = v_0 + 2\pi n$ and $v = \pi-v_0 + 2\pi n$

In your case: $\cos{x} = 0$ or $\sin{x} = -\frac{1}{4}$

**metaname90** · November 27th 2010, 08:08 PM

When does sinx=-1/4... I can't find a special triangle or unit circle or graph to solve it...

**HallsofIvy** · November 28th 2010, 03:56 AM

Originally Posted by metaname90

Okay, I always struggle with how to exactly work out the algebra for these trigonometric problems... The question is to identify where and what type of critical points lie on the function:

$f(x)=cos(2x)-1sin(x)$

Is that really what the problem said? That "1" in front of the sin(x) doesn't add anything and is peculiar.

between 0 and 2pi. Apparently there are 4.

So far this is what I have done:
$f'(x)=-2sin(2x)-cosx$

$0=-4sinxcosx-cosx$

$cosx=-4sinxcosx$

$-4sinx=1$

If $cos(x)\ne 0$

$sinx=-1/4$

First of all I don't know how to solve for x in the final line without a calculator. Also, I only see that that would only possibly lead to two critical points between 0 and 2pi....

Thank you anyone who can help me a bit!

P.S. I kind of suck at this math formatting so forgive me...

As Mondreus told you, either cos(x)= 0, which happens at $x= \frac{\pi}{2}$ and at $x= \frac{3\pi}{2}$ or sin(x)= -1/4 which happens at x= 3.3943 and at x= 6.0305, approximately. There is no "simple" form for that. Why do you want to solve it "without a calculator"?

**skeeter** · November 28th 2010, 04:13 AM

$0 = -4\sin{x}\cos{x} - \cos{x}$

factoring ensures you do not miss possible solutions ...

$0 = -\cos{x}(4\sin{x} + 1)$

$\cos{x} = 0$ , $\sin{x} = -\frac{1}{4}$

for $0 \le x < 2\pi$ ...

$x = \frac{\pi}{2}$ , $x = \frac{3\pi}{2}$

$x = 2\pi + \arcsin\left(-\frac{1}{4}\right)$

$x = \pi - \arcsin\left(-\frac{1}{4}\right)$

**metaname90** · November 28th 2010, 08:37 AM

I got the solutions now.... and yeah, the best I could do for sinx=-1/4 was 2pi+arcsin(-1/4) so I was wondering if I was missing out. In my calculus calculators are not permitted, in any of the quizzes or final exam, so I thought this was a strange solution.

And whoever pointed out the "1" in front of sinx I know that is very peculiar, I don't know why I decided to copy it directly from the question, but that's definitely how it appeared.

I found a video and then skeeter explained that you must factor in order to not lose answers, so that's all good.

Thank you everyone!

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