# Thread: How do I solve this equation?

1. ## How do I solve this equation?

Could somebody help me to solve this equation. It's been a loooooong time since highschool.

(1/sinx) + (1/cosx) = 10

x= ??? degrees

2. ## Re: How do I solve this equation?

Maybe start by mulitpling through by $\sin x$

3. ## Re: How do I solve this equation?

Originally Posted by taco120
Could somebody help me to solve this equation. It's been a loooooong time since highschool.

(1/sinx) + (1/cosx) = 10

x= ??? degrees
You have to solve following system of equations :

$\begin{cases}\cos x + \sin x=10\cdot \sin x\cdot \cos x \\\sin^2 x+\cos^2 x=1\end{cases}$

4. ## Re: How do I solve this equation?

Thanks for the replies. I think I solved it. Anyone cares to doublecheck?

$\frac{1}{sin\alpha}+\frac{1}{cos\alpha}=10 \Rightarrow$

$\frac{sin\alpha+cos\alpha}{sin\alpha \times cos\alpha}=10 \Rightarrow$

$sin\alpha+cos\alpha=10 \times sin\alpha \times cos\alpha \Rightarrow$

Identity: $sin(2\alpha)= 2 \times sin\alpha \times cos\alpha$

Times 5: $5 \times sin(2\alpha)=10 \times sin\alpha \times cos\alpha \Rightarrow$

$5 \times sin(2\alpha)=sin\alpha+cos\alpha$

Squaring: $25 \times sin^2(2\alpha)=sin^2\alpha+cos^2\alpha+2 \times sin\alpha \times cos\alpha$

Identity: $sin^2\alpha+cos^2\alpha=1$

$\Rightarrow 25 \times sin^2(2\alpha)=1+2 \times sin\alpha \times cos\alpha$

$\Rightarrow 25 \times sin^2(2\alpha)=1+sin(2\alpha)$ (following first identity)

$\Rightarrow 25 \times sin^2(2\alpha)-sin(2\alpha)-1=0$

$x=sin(2\alpha)$

$\Rightarrow 25x^2 -x -1 =0$

We now have a quadratic equation, which can be solved with the quadratic formula.

5. ## Re: How do I solve this equation?

Looks good, the solutions will be disgusting though!

6. ## Re: How do I solve this equation?

Note That:There are more than 2 solutions to this equation.

$\frac{1}{\sin{x}}+\frac{1}{\cos{x}}=10$

Express the equation in terms of $\cos{x}$
$100\cos^4{x}-20\cos^3{x}-98\cos^2{x}+20\cos{x}-1=0$

Factorizing this (with the help of wolfram alpha):

\begin{align*} (\cos{x}-0.993799)(\cos{x}-0.111188)(\cos{x}-0.0908748)(\cos{x}+0.995862)=1 \end{align*}

*These are not accurate.

We obtain $\cos{x} \approx 0.993799, \hspace{4} 0.111188, \hspace{4} 0.0908748, \hspace{4} -0.995862$