\[\text{trig}_a(\theta) + \text{trig}_b(2\theta) = 4\] \[\text{Solve for } \theta \] \[\text{where;}\qquad \{\text{ trig}_a(\theta )~\}\qquad \text{is some trigonometric function}\] \[\text{and;}\qquad \{\text{ trig}_b(2\theta )~\}\qquad \text{is some other trigonometric function}\]
Is a general solution to this equation possible? @Mr.Math @FoolForMath
well i know of a simple solution
I think this is artificially made harder by veiling the trig functions.
well at least one of them must be greater than 1
I am curious, what's the general solution?
im not looking for a general solution i think that would be too hard, im looking for a specific solution, to solve this you might need to know some trig functions that can be greater than zero
Hmm tan, cot hmm sec, csc.
getting warm...
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