$\text{If}{tan}^{-1}(-3x)+{tan}^{-1}(-2x)=\frac{\pi }{4},\text{then the values of}\mathrm{ x}\text{a re}$

The correct option is:
$1,\frac{-1}{6}$

Step-by-step Explanation:

We are given the following trigonometric equation:
${tan}^{-1}(-3x)+{tan}^{-1}(-2x)=\frac{\pi }{4}$

First, recall the property of the inverse tangent function:
${tan}^{-1}(-\theta )=-{tan}^{-1}\left(\theta \right)$

Applying this property to both terms on the left-hand side of the equation, we get:
$-{tan}^{-1}\left(3x\right)-{tan}^{-1}\left(2x\right)=\frac{\pi }{4}$

Multiplying the entire equation by -1 gives:
${tan}^{-1}\left(3x\right)+{tan}^{-1}\left(2x\right)=-\frac{\pi }{4}$

Now, we apply the tangent function to both sides of the equation:
$tan\left[{tan}^{-1}\left(3x\right)+{tan}^{-1}\left(2x\right)\right]=tan\left(-\frac{\pi }{4}\right)$

Using the sum formula for tangent, $tan(A+B)=\frac{tan\left(A\right)+tan\left(B\right)}{1-tan\left(A\right)tan\left(B\right)}$, we can rewrite the left side. Since $tan\left(-\frac{\pi }{4}\right)=-1$, we get:
$\frac{3x+2x}{1-\left(3x\right)\left(2x\right)}=-1$

Simplifying the fraction:

Multiply both sides by $1-6x^2$:
$5x=-1(1-6x^2)$

Simplify and rearrange the equation to form a quadratic equation:
$5x=-1+6x^2$
$6x^2-5x-1=0$

To solve the quadratic equation, we factorize by splitting the middle term:
$6x^2-6x+x-1=0$
$6x(x-1)+1(x-1)=0$
$(6x+1)(x-1)=0$

Equating each factor to zero yields:
$x=1\text{or}x=-\frac{1}{6}$