Trigonometry solutions Question

PhamCy
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Summary: https://www.physicsforums.com/threads/trigonometry-question.977263/

Here's the question.
Find the solutions of the equation tan(x)=2cos(x)+1 if 0 ≤ x ≤ 2π.
I know this question can be solved by observing the graph but is there any other ways (like algorithms OR some Trigonometry rules) to solve this, I couldn't figure that out.
Thank you for those who noticed my question.

[Moderator's note: Moved from a technical forum and thus no template.]
 
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Hello PhamCy, :welcome: !

Nice try, but PF asks a little more from you: make a start with the solution and we'll help you.

For example: substitute ##\tan x = {\sin x\over \cos x}## and see where it leads you
 
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Likes Greg Bernhardt
BvU said:
see where it leads you
Way beyond what you can solve analytically with high school mathematics. WolframAlpha knows how to solve it analytically, here is one of the solutions:

equation.png
 
Oops, I 'missed' the +1 which makes things difficult o:)
 
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Likes Delta2 and Klystron
You could express the relationship as a polynomial in ##\cos\theta##, and then find the roots of the polynomial using something like Newton's method.
 
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Likes Delta2
Turn everything into \cos x to get:
<br /> 4\cos^{4}x+4\cos^{3}x+2\cos^{2}x-1=0<br />
From here solve it using whatever method you like. I would advise you plotting the polynomial first.
 
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Likes Delta2
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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