# Solve for x that Satisfies the Trigonometric Equation

• Nov 9th 2011, 06:39 PM
jigogwapo16
Solve for x that Satisfies the Trigonometric Equation
Hi, do any of you know how to solve this problem?

Find the only value of x in $(-\frac{\pi}{2},0)$ that satisfies the equation

$\frac{\sqrt{3}}{\sin x} + \frac{1}{\cos x} = 4$

The answer was $-\frac{4\pi}{9}$, but I don't know how it came to that answer.

Thanks!
• Nov 9th 2011, 07:04 PM
TKHunny
Re: Solve for x that Satisfies the Trigonometric Equation
First, I'm tempted to change to csc(x) and sec(x), but like many, maybe I'm not all that comfortable with those two functions.
Second, I'm tempted to multiply by sin(x), since sin(x) is never zero in the given interval. Maybe I'm lots more familar with the tangent function.
Third, maybe I'm tempted to get rid of the cosines by the Pythagorean Identity.

Not sure right off. What have you tried?
• Nov 9th 2011, 07:18 PM
Deveno
Re: Solve for x that Satisfies the Trigonometric Equation
i think fractions are evil. i try to get rid of them.
• Nov 9th 2011, 09:47 PM
jigogwapo16
Re: Solve for x that Satisfies the Trigonometric Equation
I'm actually pretty stumped in this problem. This was actually a problem in the national round of one of the math contests here in our country.

The first thing I did was that I ignored the interval $(-\frac{\pi}{2},0)$ I would have $x = \frac{\pi}{3}$ as the answer, since $\sin x = \frac{\sqrt{3}}{2}$ and $\cos x = \frac{1}{2}$. But then I saw the friggin' interval restriction, and I got lost.
• Nov 9th 2011, 10:08 PM
jigogwapo16
Re: Solve for x that Satisfies the Trigonometric Equation
Quote:

Originally Posted by TKHunny
First, I'm tempted to change to csc(x) and sec(x), but like many, maybe I'm not all that comfortable with those two functions.
Second, I'm tempted to multiply by sin(x), since sin(x) is never zero in the given interval. Maybe I'm lots more familar with the tangent function.
Third, maybe I'm tempted to get rid of the cosines by the Pythagorean Identity.

Not sure right off. What have you tried?

I tried multiplying everything by sin(x)cos(x), but I was still stuck since I'm left with both sin x and cos x.

I also tried squaring both sides, but I still have the sinxcosx to deal with when squaring the binomial in the LHS of the equation.
• Nov 10th 2011, 12:49 AM
sbhatnagar
Re: Solve for x that Satisfies the Trigonometric Equation
$\frac{\sqrt{3}}{\sin{x}}+\frac{1}{\cos{x}}=4$

Quote:

Expressing the equation in terms of cos(x):
$\sin{x}=\frac{\sqrt{3}\cos{x}}{4\cos{x}-1}$

$1-\cos^2{x}=(\frac{\sqrt{3}\cos{x}}{4\cos{x}-1})^2$

$16\cos^4{x}-8\cos^3{x}-12\cos^2{x}+8\cos{x}-1=0$.

Quote:

Let $u=\cos{x}$. Equation becomes:
$16u^4-8u^3-12u^2+8u-1=0$

Quote:

Factorising:
$(2u-1)(8u^3-6u+1)=0$

$\\2u-1=0\\\\u=\frac{1}{2} \\\\ \cos{x}=\frac{1}{2} \\\\ x= \frac{1}{3}(6\pi n \pm \pi) \hspace{6} , n \in \mathbb{Z}$.

Quote:

$\text{Note That:}$This is not the only solution to x. Let us also solve $8u^3-6u+1=0$.
$8u^3-6u+1=0$

$8\cos^3{x}-6\cos{x}+1=0$

$2\cos{3x}+1=0$

$\cos{3x}=-\frac{1}{2}$

$x=\frac{2}{9}(3\pi n \pm \pi) \hspace{6} , n \in \mathbb{Z}$

Quote:

$x= \frac{1}{3}(6\pi n \pm \pi) , \hspace{6} \frac{2}{9}(3\pi n \pm \pi) \hspace{6} , n \in \mathbb{Z}$

Quote:

Originally Posted by jigogwapo16
Hi, do any of you know how to solve this problem?

Find the only value of x in $(-\frac{\pi}{2},0)$ that satisfies the equation

$\frac{\sqrt{3}}{\sin x} + \frac{1}{\cos x} = 4$

The answer was $-\frac{4\pi}{9}$, but I don't know how it came to that answer.

Thanks!

Can you find the value of x in $[-\frac{\pi}{2},0]$ that satisfies the equation using the general solution from the second factor?

$x=\frac{2}{9}(3\pi n \pm \pi) \hspace{6} , n \in \mathbb{Z}$
• Nov 10th 2011, 07:13 PM
jigogwapo16
Re: Solve for x that Satisfies the Trigonometric Equation
Quote:

Originally Posted by sbhatnagar
$\frac{\sqrt{3}}{\sin{x}}+\frac{1}{\cos{x}}=4$

$\sin{x}=\frac{\sqrt{3}\cos{x}}{4\cos{x}-1}$

$1-\cos^2{x}=(\frac{\sqrt{3}\cos{x}}{4\cos{x}-1})^2$

$16\cos^4{x}-8\cos^3{x}-12\cos^2{x}+8\cos{x}-1=0$.

$16u^4-8u^3-12u^2+8u-1=0$

$(2u-1)(8u^3-6u+1)=0$

$\\2u-1=0\\\\u=\frac{1}{2} \\\\ \cos{x}=\frac{1}{2} \\\\ x= \frac{1}{3}(6\pi n \pm \pi) \hspace{6} , n \in \mathbb{Z}$.

$8u^3-6u+1=0$

$8\cos^3{x}-6\cos{x}+1=0$

$2\cos{3x}+1=0$

$\cos{3x}=-\frac{1}{2}$

$x=\frac{2}{9}(3\pi n \pm \pi) \hspace{6} , n \in \mathbb{Z}$

$x= \frac{1}{3}(6\pi n \pm \pi) , \hspace{6} \frac{2}{9}(3\pi n \pm \pi) \hspace{6} , n \in \mathbb{Z}$

Can you find the value of x in $[-\frac{\pi}{2},0]$ that satisfies the equation using the general solution from the second factor?

$x=\frac{2}{9}(3\pi n \pm \pi) \hspace{6} , n \in \mathbb{Z}$

Thank you very much! You're a lifesaver! :) I actually thought of expressing everything in cos x, but the steps I took were a little more complicated. I reached an order 8 equation, but still couldn't get rid of sin x. Thanks again!