Answer:
See below
Step-by-step explanation:
It has something to do with the Weierstrass substitution, where we have
[tex]$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$[/tex]
First, consider the double angle formula for tangent:
[tex]\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}[/tex]
Therefore,
[tex]\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}[/tex]
Once the double angle identity for sine is
[tex]\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}[/tex]
we know [tex]\sin(x)=\dfrac{2t}{1+t^2}[/tex], but sure, we can derive this formula considering the double angle identity
[tex]\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)[/tex]
Recall
[tex]\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}[/tex]
Thus,
[tex]\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}[/tex]
Similarly for cosine, consider the double angle identity
Thus,
[tex]\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}[/tex]
Hence, we showed [tex]\sin(x) \text { and } \cos(x)[/tex]
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[tex]5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ][/tex]
Solving
[tex]5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3[/tex]
[tex]\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3[/tex]
[tex]\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0[/tex]
[tex]t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}[/tex]
[tex]t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}[/tex]
Just note that
[tex]\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}[/tex]
and [tex]\tan\left(\dfrac{x}{2}\right)[/tex] is not defined for [tex]x=k\pi , k\in\mathbb{Z}[/tex]
