Nathalie Fields

2022-07-25

Solve the equation on the interval $\left[0,2\pi \right]$
$\mathrm{tan}2x-\mathrm{tan}x=0$

Izabelle Frost

Expert

We then divide $\mathrm{tan}2x$ and $\mathrm{tan}x$ by what has been taken out or $\mathrm{tan}x$.
After factoring, we get this trig equation in factored form:
$\mathrm{tan}x\left(\mathrm{tan}x-1\right)=0$
We have two factors and they are:
$\mathrm{tan}x$ and $\mathrm{tan}x-1$
Set each factor to zero just like you did in algebra 1 andsolve for the given trig function.
$\mathrm{tan}x=0$
When does $\mathrm{tan}x=0$?
$\mathrm{tan}x=0$ at 0 degrees (the same as 360 degrees) and 180 degrees.
We now set the other factor to 0 and solve the equation.
$\mathrm{tan}x-1=0$
Adding 1 to both sides we get:
$\mathrm{tan}x=1$
When does $\mathrm{tan}x=1$?
$\mathrm{tan}x=1$ at 45 degrees and 225 degrees
All degrees between the given interval are: 0 degrees (or 360degrees), 180 degrees, 45 degrees and 225 degrees.
$\frac{\pi }{4},\pi ,\frac{5\pi }{4}$, and $2\pi$

valtricotinevh

Expert

$\mathrm{tan}2x=\frac{2\mathrm{tan}\left(x\right)}{1-{\mathrm{tan}}^{2}\left(x\right)}$
$\frac{2\mathrm{tan}\left(x\right)}{1-{\mathrm{tan}}^{2}\left(x\right)}-\frac{\mathrm{tan}\left(x\right)}{1-{\mathrm{tan}}^{2}\left(x\right)}=0$
$\frac{2\mathrm{tan}\left(x\right)-\mathrm{tan}\left(x\right)\left(1-{\mathrm{tan}}^{2}\left(x\right)\right)}{1-{\mathrm{tan}}^{2}\left(x\right)}=0$
$2\mathrm{tan}\left(x\right)-\mathrm{tan}\left(x\right)+\mathrm{tan}\left(x\right){\mathrm{tan}}^{2}\left(x\right)=0$
$\mathrm{tan}\left(x\right)+\mathrm{tan}\left(x\right){\mathrm{tan}}^{2}\left(x\right)=0$
$\mathrm{tan}\left(x\right)\left(1+{\mathrm{tan}}^{2}\left(x\right)\right)=0$
$\mathrm{tan}\left(x\right)=0$ or ${\mathrm{tan}}^{2}\left(x\right)=-1$
x = 0 or $2\pi$

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