For what values of $x$ $x$ does $tan (- 2 x) = cot (x)$ $tan(-2x) = cot(x)$ ?

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Answer 1 · 2017-04-08T17:47:24

$x = \frac{(2 k + 1) π}{2}$ $x = ((2k+1)pi)/2" "$ for any integer $k$ $k$ .

Given:

$tan (- 2 x) = cot (x)$ $tan(-2x) = cot(x)$

That is:

$\frac{sin (- 2 x)}{cos (- 2 x)} = \frac{cos (x)}{sin (x)}$ $sin(-2x)/cos(-2x) = cos(x)/sin(x)$

Multiplying both sides by $cos (- 2 x) sin (x)$ $cos(-2x)sin(x)$ , this becomes:

$sin (x) sin (- 2 x) = cos (x) cos (- 2 x)$ $sin(x)sin(-2x) = cos(x)cos(-2x)$

Subtracting $sin (- 2 x) sin (x)$ $sin(-2x)sin(x)$ from both sides, we get:

$cos (x) cos (- 2 x) - sin (x) sin (- 2 x) = 0$ $cos(x)cos(-2x)-sin(x)sin(-2x) = 0$

Compare the left hand side with the sum formula for $cos$ $cos$ :

$cos (α) cos (β) - sin (α) sin (β) = cos (α + β)$ $cos(alpha)cos(beta)-sin(alpha)sin(beta) = cos(alpha+beta)$

So with $α = x$ $alpha=x$ and $β = - 2 x$ $beta=-2x$ we get:

$cos (- x) = 0$ $cos(-x) = 0$

Note that $cos (- x) = cos (x)$ $cos(-x) = cos(x)$

Hence:

$x = \frac{(2 k + 1) π}{2}$ $x = ((2k+1)pi)/2$

where $k$ $k$ is any integer.

Here are the two functions $tan (- 2 x)$ $tan(-2x)$ and $cot (x)$ $cot(x)$ plotted together:

graph{(y+tan(2x))(y-cot(x)) = 0 [-10, 10, -5, 5]}

For what values of xxx does tan(-2x) = cot(x)tan(−2x)=cot(x)tan(-2x) = cot(x) ?