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

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

x= ??? degrees

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- Nov 28th 2011, 11:15 AMtaco120How 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 - Nov 28th 2011, 11:51 AMpickslidesRe: How do I solve this equation?
Maybe start by mulitpling through by $\displaystyle \sin x$

- Nov 29th 2011, 07:46 AMprincepsRe: How do I solve this equation?
- Nov 29th 2011, 12:08 PMtaco120Re: How do I solve this equation?
Thanks for the replies. I think I solved it. Anyone cares to doublecheck?

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle x=sin(2\alpha)$

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

We now have a quadratic equation, which can be solved with the quadratic formula. - Nov 29th 2011, 12:29 PMQuackyRe: How do I solve this equation?
Looks good, the solutions will be disgusting though!

- Dec 1st 2011, 03:45 AMsbhatnagarRe: How do I solve this equation?
**Note That**:There are more than 2 solutions to this equation.

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

Quote:

Express the equation in terms of $\displaystyle \cos{x}$

Factorizing this (with the help of wolfram alpha):

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

*These are not accurate.

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