Gregory Jones

2021-12-28

Finding solution of a Trigonometric equation
$\mathrm{tan}A+\mathrm{tan}2A+\mathrm{tan}3A=0$
I tried converting these all in sin and cos and I got the answer but the answer didn't match

Hattie Schaeffer

Use this:
$\mathrm{tan}3A=\frac{\mathrm{tan}A+\mathrm{tan}2A}{1-\mathrm{tan}A\mathrm{tan}2A}$
Substitute the numerator with −$\mathrm{tan}3A$ and youre

enhebrevz

$\mathrm{tan}A+\mathrm{tan}2A=\frac{\mathrm{sin}3A}{\mathrm{cos}A\mathrm{cos}2A}$
$\mathrm{tan}A+\mathrm{tan}2A+\mathrm{tan}3A=0⇒\frac{\mathrm{sin}3A}{\mathrm{cos}A\mathrm{cos}2A}+\frac{\mathrm{sin}3A}{\mathrm{cos}3A}=0$
$⇔\mathrm{sin}3A\left(\mathrm{cos}3A+\mathrm{cos}A\mathrm{cos}2A\right)=0$
If where n is any integer
Otherwise,
$\mathrm{cos}3A+\mathrm{cos}A\mathrm{cos}2A=0⇔4\mathrm{cos}3A-3\mathrm{cos}A+\mathrm{cos}A\left(2{\mathrm{cos}}^{2}A-1\right)=0$
$⇔\mathrm{cos}A\left(6{\mathrm{cos}}^{2}A-4\right)=0$
If
Otherwise,

nick1337

Hint
The following identities for the tangent of multiple angles should be useful
$\mathrm{tan}2A=\frac{2\mathrm{tan}A}{1-2{\mathrm{tan}}^{2}A}$
$\mathrm{tan}3A=\frac{3\mathrm{tan}A-{\mathrm{tan}}^{3}A}{1-3{\mathrm{tan}}^{2}A}$
I am sure that you can take from here.

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