How do you solve $2 {sin}^{2} x tan x = tan x$ $2sin^2xtanx=tanx$ ?

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Answer 1 · 2016-09-22T12:12:06

$x = 0, \frac{π}{4}$ $x = 0, (pi) / (4)$

We have: $2 {sin}^{2} (x) tan (x) = tan (x)$ $2 sin^(2)(x) tan(x) = tan(x)$

$\Rightarrow 2 {sin}^{2} (x) tan (x) - tan (x) = 0$ $=> 2 sin^(2)(x) tan(x) - tan(x) = 0$

$\Rightarrow tan (x) (2 {sin}^{2} (x) - 1) = 0$ $=> tan(x) (2 sin^(2)(x) - 1) = 0$

$\Rightarrow tan (x) = 0$ $=> tan(x) = 0$

$\Rightarrow x = arctan (0)$ $=> x = arctan(0)$

$\Rightarrow x = 0$ $=> x = 0$

or

$\Rightarrow 2 {sin}^{2} (x) - 1 = 0$ $=> 2 sin^(2)(x) - 1 = 0$

$\Rightarrow {sin}^{2} (x) = \frac{1}{2}$ $=> sin^(2)(x) = (1) / (2)$

$\Rightarrow sin (x) = \frac{\sqrt{2}}{2}$ $=> sin(x) = (sqrt(2)) / (2)$

$\Rightarrow x = arcsin (\frac{\sqrt{2}}{2})$ $=> x = arcsin((sqrt(2)) / (2))$

$\Rightarrow x = \frac{π}{4}$ $=> x = (pi) / (4)$

Therefore, the solutions to the equation are $x = 0$ $x = 0$ and $x = \frac{π}{4}$ $x = (pi) / (4)$ .

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