Find the solution of the equation if $\displaystyle 0 \leq t \leq 2\pi $

Question

$\displaystyle tan^2t-sect=1 $

Attempt

$\displaystyle tan^2-\frac{1}{tant}-1=0 $

I have no idea what to do next?

Thank you

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- Mar 24th 2009, 10:40 AMmj.alawami[SOLVED] Solving Trigonometric equations 2
Find the solution of the equation if $\displaystyle 0 \leq t \leq 2\pi $

**Question**

$\displaystyle tan^2t-sect=1 $

**Attempt**

$\displaystyle tan^2-\frac{1}{tant}-1=0 $

I have no idea what to do next?

Thank you - Mar 24th 2009, 01:32 PMOpalg
That's a reasonable attempt, but the problem is that if you multiply through by tan(t) then you'll have a cubic equation in tan(t), which doesn't look promising.

A better strategy would be to use the fact that $\displaystyle \tan^2t = \sec^2t-1$. Then the original problem becomes $\displaystyle \sec^2t - \sec t -2 = 0$. That's a quadratic in sec(t), which you should be able to solve.