cos(x)/sin(x)+1/cos(x)=5.
Let c=cos(x), then:
c/√(1-c2)=5-1/c.
Square both sides:
c2/(1-c2)=25-10/c+1/c2.
Multiply through by c2(1-c2):
c4=25c2(1-c2)-10c(1-c2)+1-c2=25c2-25c4-10c+10c3+1-c2,
26c4-10c3-24c2+10c-1=0, but this doesn't factorise into rational factors, but c=±1 would have been roots if the constant had been -2 instead of -1. This tells us that the true roots include numbers close to 1 and -1. We can use this fact to solve the quartic equation.
This method requires calculus: let f(c)=26c4-10c3-24c2+10c-1. We need the derivative f'(c)=104c3-30c2-48c+10. Next we use Newton's iterative method for finding solutions:
cn+1=cn-f(cn)/f'(cn).
We need a starting value, c0, so let c0=1 initially, then c1=1-f(1)/f'(1)=1-1/36=35/36.
c2=0.9697..., c3=0.96969..., c4=0.9696916..., c5=0.9696916143..., c6=0.969691614318.
Now put c0=-1, then c=-0.986454731614.
We must now test both these values to ensure that they satisfy the original equation.
When c=0.969691614318, x=0.24683 radians (approx). This value satisfies the original equation.
When c=-0.986454731614, x=2.97681 radians (approx). This value doesn't satisfy the original equation.
A graph of f(x) reveals another solution for c. This value is close to zero, so let c0=0. c1=1/10. Continuing the iterations we get c=0.192451383033.
Therefore x=1.37714 radians (approx). This value doesn't satisfy the original equation, but comes fairly close to 5, suggesting that there may be another root close by.
The remaining root is c=0.20893, making x=1.36032. This value satisfies the original equation. Therefore we have two solutions for cos(x)=0.96969 or 0.20893 (approx).
There are infinitely more solutions for x because of the cyclic nature of the cosine function.