By means of graph, solve the inequality
tan2x+1≤4
I’ve tried like (2tanx)/(1-tan^{2}) -3≤0
then (3tan^{2}x+2tanx-3)/(1+tanx)(1-tanx) ≤0
…
can anyone point out my mistake?
If you have properly described the question, you are to solve this BY MEANS OF A GRAPH.
As topsquark pointed out, this requires sketching the pertinent graph and no analysis at all.
To deal with this analytically rather than geometrically, I'd probably start with u = 2x which reduces the problem to $tan(u) \le 3.$
Then I'd use the inverse trig functions.
How you answer depends on exactly how the question is asked and what you have been taught.
Are you to answer in radians or degrees? Are there any additional limits placed on x? Do you know how to represent a cyclic answer?
A cyclic answer in radians looks like this:
$cos( \theta ) = 0 \implies \theta = \dfrac{\pi }{2} * (2k - 1)\ for\ all\ k \in \mathbb Z.$
[QUOTE=JeffM;823365]How you answer depends on exactly how the question is asked and what you have been taught.
Are you to answer in radians or degrees? Are there any additional limits placed on x? Do you know how to represent a cyclic answer?
answer in degree. And there's no additional limits
$tan(2x) + 1 \le 4 \implies tan(2x) \le 3.$
$0 < v < 90^o\ and\ v = arctan(3) \implies v \approx 71.5650^o < 90^o.$
$-90^o < 2x < 90^o\ and\ tan(2x) \le 3 \implies -45^o < x\ and\ 2x \le v \implies - 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$
So one possible answer is $- 45^o < x \le \dfrac{arcsin(3)}{2} \approx 35.7825^o.$
A better answer might be $\{-45 + (2k - 1) * 90\}^o < x \le \left\{\dfrac{arcsin(3)}{2} + (2k - 1) * 90\right\}^o,\ for\ all\ k \in \mathbb Z.$