Number of solutions of the equation
tan2x=2cos(x+π) in
(−π,π) are same as number of points of intersection of
y=tan2x and
y=2cos(x+π) in
(−π,π).
Here fundamental functions involved are
tanx and
cosx.
Whose graphs are given by
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1419334/original_sinx.png)
and
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1419341/original_sinx.png)
Now apply the stretch transformation by 2 units to
y=tanx to get the graph of
y=tan2x as shown below
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1419345/original_sinx.png)
Now we will make transformation to the graph of
y=cosx.
First, apply the horizontal shift by
π units to get
y=cos(x−π) as shown below
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1419349/original_sinx.png)
Further, apply stretch transformation along y by 2 units to get
y=2cos(x+π) as shown below
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1419353/original_sinx.png)
Now we will calculate the number of points of intersection in combined graph of
y=tan2x and
y=2cos(x+π)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1419357/original_sinx.png)
From the above graph, we can see there are 4 points of intersection in
(−π,π).
Hence the given equation have 4 solutions in
(−π,π)