Selena Cowan

2022-01-23

what is x for ${\mathrm{tan}}^{2}3x=2{\mathrm{sin}}^{2}3x$

trasahed

Expert

${\mathrm{tan}}^{2}3x=2{\mathrm{sin}}^{2}3x$
${\mathrm{tan}}^{2}3x-2{\mathrm{sin}}^{2}3x=0$
${\mathrm{sin}}^{2}3x\left(\frac{1}{{\mathrm{cos}}^{2}3x}-2\right)=0$
${\mathrm{sin}}^{2}3x\left({\mathrm{sec}}^{2}3x-2\right)=0$
So we have either
${\mathrm{sin}}^{2}3x=0$
$3x={0}^{\circ },{180}^{\circ },{360}^{\circ }$
$x={0}^{\circ },{60}^{\circ },{120}^{\circ }$
or
${\mathrm{sec}}^{2}3x-2=0$
${\mathrm{sec}}^{2}3x=2$
$\mathrm{sec}3x=±\sqrt{2}$
$3x={45}^{\circ },{135}^{\circ },{225}^{\circ },{315}^{\circ }$
$x={15}^{\circ },{45}^{\circ },{75}^{\circ },{105}^{\circ }$
From the above, the only admissible solutions are ${15}^{\circ },{45}^{\circ },{60}^{\circ },{75}^{\circ }$
The sum of these angles is ${195}^{\circ }$

Allison Compton

Expert

Do some trigonometry and find the general solution first:
${\mathrm{tan}}^{2}3x=2{\mathrm{sin}}^{2}3x=\frac{2{\mathrm{tan}}^{2}3x}{1+{\mathrm{tan}}^{2}3x}⇔{\mathrm{tan}}^{2}3x\left(1+{\mathrm{tan}}^{2}3x\right)=2{\mathrm{tan}}^{2}3x$
So,
either ;
or
Then select the values in [0,90]

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