Solve tan2(x)−(1+3)tan⁡(x)+3&lt;0But I get D=4−23 and I don't think that's the right way to go about this.

"Solve tan2(x)−(1+3)tan⁡(x)+3<0But I get D=4−23 and I don't..." was answered by students like you

ropowiec2gkc

2022-03-21

Solve

\tan^{2} (x) - (1 + \sqrt{3}) \tan (x) + \sqrt{3} < 0

But I get

D = 4 - 2 \sqrt{3}

and I don't think that's the right way to go about this.

cineworld93uowb

Beginner2022-03-22Added 16 answers

Explanation:
How about changing the question?

(\tan x - 1) (\tan x - \sqrt{3}) < 0

So the zeroes of that function are

\frac{π}{4} and \frac{π}{3}

.
If you look at the values in brackets, both are negative for

x < \frac{π}{4}

, meaning the final value is positive. For

x > \frac{π}{3}

both values in brackets are positive, meaning the final value is also positive.
Let's analyse that interval

x \in (d \frac{π}{4}, d \frac{π}{3})

. In this interval,

(\tan x - 1)

is positive but

(\tan x - \sqrt{3})

is negative, meaning the final result is negative and hence

< 0

.
So here's your final answer:

d \frac{π}{4} + 2 π n < x < d \frac{π}{3} + 2 π n n \in N

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