If 2\tan^2x - 5\sec x = 1 has exactly 7 distinct solutions for x\in[0,\frac{n\pi}{2}],\ \

Jakayla Benton

Jakayla Benton

Answered question

2022-04-23

If 2tan2x5secx=1 has exactly 7 distinct solutions for x[0,nπ2],  nN, then the greatest value of n is?
My attempt:
Solving the above quadratic equation, we get cosx=13
The general solution of the equation is given by cosx=2nπ±cos113
For having 7 distinct solutions, n can have value = 0,1,2,3
So, from here we can conclude that n is anything but greater than 6. So, according to the options given in the questions, the greatest value of n should be 13. But the answer given is 14. Can anyone justify?

Answer & Explanation

Addison Zamora

Addison Zamora

Beginner2022-04-24Added 10 answers

I agree with your answer, indeed note that
2tan2x5secx=12(1cos2x)5cosx=cos2x
3cos2x+5cosx2=0
and
3t2+5t2=0t=5±25+246t=5+476=13
thus we have 2 solution on the interval [0,2π] and notably one in the interval [0,π2] and the other in [3π2,2π]
Therefore, since the function is periodic with period 2π we have that n=13.
Note that for x[0,13π2] the expression is not defined when x=π2+kπ thus this points should be excluded by the solution.

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