 dream13rxs

2022-07-08

How to find the solution of this trigonometric equation
$\mathrm{cot}\left(x+{110}^{\circ }\right)=\mathrm{cot}\left(x+{60}^{\circ }\right)\mathrm{cot}x\mathrm{cot}\left(x-{60}^{\circ }\right)$
I have used the formulae
$\mathrm{cos}\left(x+{60}^{\circ }\right)\mathrm{cos}\left(x-{60}^{\circ }\right)={\mathrm{cos}}^{2}{60}^{\circ }-{\mathrm{sin}}^{2}x$
$\mathrm{sin}\left(x+{60}^{\circ }\right)\mathrm{sin}\left(x-{60}^{\circ }\right)={\mathrm{sin}}^{2}x-{\mathrm{sin}}^{2}{60}^{\circ }$
How to move further? What is the least positive value of x? Brendan Bush

Expert

$\mathrm{cot}\left(x+{110}^{\circ }\right)=\mathrm{cot}\left(x+{60}^{\circ }\right)\mathrm{cot}x\mathrm{cot}\left(x-{60}^{\circ }\right)$
$\frac{\mathrm{cot}\left(x+{110}^{\circ }\right)}{\mathrm{cot}x}=\mathrm{cot}\left(x+{60}^{\circ }\right)\cdot \mathrm{cot}\left(x-{60}^{\circ }\right)$
$\frac{\mathrm{cos}\left(x+{110}^{\circ }\right)\cdot \mathrm{sin}x}{\mathrm{sin}\left(x+{110}^{\circ }\right)\cdot \mathrm{cos}x}=\frac{\mathrm{cos}\left(x+{60}^{\circ }\right)\cdot \mathrm{cos}\left(x-{60}^{\circ }\right)}{\mathrm{sin}\left(x+{60}^{\circ }\right)\cdot \mathrm{sin}\left(x-{60}^{\circ }\right)}$
Now Using Componendo and Dividendo, We get
$\frac{\mathrm{cos}\left(x+{110}^{\circ }\right)\cdot \mathrm{sin}x+\mathrm{sin}\left(x+{110}^{\circ }\right)\cdot \mathrm{cos}x}{\mathrm{cos}\left(x+{110}^{\circ }\right)\cdot \mathrm{sin}x-\mathrm{sin}\left(x+{110}^{\circ }\right)\cdot \mathrm{cos}x}=\frac{\mathrm{cos}\left(x+{60}^{\circ }\right)\cdot \mathrm{cos}\left(x-{60}^{\circ }\right)+\mathrm{sin}\left(x+{60}^{\circ }\right)\cdot \mathrm{sin}\left(x-{60}^{\circ }\right)}{\mathrm{cos}\left(x+{60}^{\circ }\right)\cdot \mathrm{cos}\left(x-{60}^{\circ }\right)-\mathrm{sin}\left(x+{60}^{\circ }\right)\cdot \mathrm{sin}\left(x-{60}^{\circ }\right)}$
$-\frac{\mathrm{sin}\left[\left(x+{110}^{\circ }\right)+x\right]}{\mathrm{sin}\left[\left(x+{110}^{\circ }\right)-x\right]}=\frac{\mathrm{cos}\left[\left(x+{60}^{\circ }\right)-\left(x-{60}^{\circ }\right)\right]}{\mathrm{cos}\left[\left(x+{60}^{\circ }\right)+\left(x-{60}^{\circ }\right)\right]}$
So we get
$-\frac{\mathrm{sin}\left(2x+{110}^{\circ }\right)}{\mathrm{sin}\left({110}^{\circ }\right)}=\frac{\mathrm{cos}\left({120}^{\circ }\right)}{\mathrm{cos}\left(2x\right)}$
$\mathrm{sin}\left(2x+{110}^{\circ }\right)\cdot \mathrm{cos}\left(2x\right)=\frac{1}{2}\mathrm{sin}\left({110}^{0}\right)$
So we get
$2\mathrm{sin}\left(2x+{110}^{\circ }\right)\cdot \mathrm{cos}\left(2x\right)=\mathrm{sin}\left({110}^{0}\right)$
$\mathrm{sin}\left(4x+{110}^{\circ }\right)+\mathrm{sin}\left({110}^{\circ }\right)=\mathrm{sin}\left({110}^{\circ }\right)$
So we get
$\mathrm{sin}\left(4x+{110}^{\circ }\right)=0⇒4x+{110}^{0}=n×{180}^{\circ }\phantom{\rule{thickmathspace}{0ex}},$
Where $n\in \mathbb{Z}$

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