if cot(x)+sec(x)=5, then cos(x)=?

Question

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if cot(x)+sec(x)=5, then cos(x)=?

Answer 1

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:

c_n+1=c_n-f(c_n)/f'(c_n).

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.