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 $\displaystyle (-\frac{\pi}{2},0)$ that satisfies the equation

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

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

Thanks!

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?

Re: Solve for x that Satisfies the Trigonometric Equation

i think fractions are evil. i try to get rid of them.

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 $\displaystyle (-\frac{\pi}{2},0)$ I would have $\displaystyle x = \frac{\pi}{3}$ as the answer, since $\displaystyle \sin x = \frac{\sqrt{3}}{2}$ and $\displaystyle \cos x = \frac{1}{2}$. But then I saw the friggin' interval restriction, and I got lost.

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.

Re: Solve for x that Satisfies the Trigonometric Equation

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

Quote:

Expressing the equation in terms of cos(x):

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

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

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

Quote:

Let $\displaystyle u=\cos{x}$. Equation becomes:

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

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

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

Quote:

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

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

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

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

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

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

Quote:

Let us finalize our answer:

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

Quote:

Originally Posted by

**jigogwapo16** Hi, do any of you know how to solve this problem?

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

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

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

Thanks!

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

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

Re: Solve for x that Satisfies the Trigonometric Equation

Quote:

Originally Posted by

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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!