Could somebody help me to solve this equation. It's been a loooooong time since highschool.
(1/sinx) + (1/cosx) = 10
x= ??? degrees
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 $
Identity: $\displaystyle sin(2\alpha)= 2 \times sin\alpha \times cos\alpha$
Times 5: $\displaystyle 5 \times sin(2\alpha)=10 \times sin\alpha \times cos\alpha \Rightarrow$
$\displaystyle 5 \times sin(2\alpha)=sin\alpha+cos\alpha$
Squaring: $\displaystyle 25 \times sin^2(2\alpha)=sin^2\alpha+cos^2\alpha+2 \times sin\alpha \times cos\alpha$
Identity: $\displaystyle sin^2\alpha+cos^2\alpha=1 $
$\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.
Note That:There are more than 2 solutions to this equation.
$\displaystyle \frac{1}{\sin{x}}+\frac{1}{\cos{x}}=10$
$\displaystyle 100\cos^4{x}-20\cos^3{x}-98\cos^2{x}+20\cos{x}-1=0$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 $\displaystyle \cos{x} \approx 0.993799, \hspace{4} 0.111188, \hspace{4} 0.0908748, \hspace{4} -0.995862$