The solution of the equation $\cos ^2 \theta +\sin \theta +1=0$, lies in the interval

a)(-$\pi$/4, $\pi$/4)
b)($\pi$/4, 3$\pi$/4)
c)(3$\pi$/4, 5$\pi$/4)
d)(5$\pi$/4, 7$\pi$/4)

Question

The solution of the equation $\cos  ^2 \theta +\sin  \theta +1=0$, lies in the interval

a)(-$\pi$/4, $\pi$/4)
b)($\pi$/4, 3$\pi$/4)
c)(3$\pi$/4, 5$\pi$/4)
d)(5$\pi$/4, 7$\pi$/4)

anonymous · Answer

cos^2 + sin^2 = 1

ujjwal · Answer

oh, I wrote the question wrong!

ujjwal · Answer

I have corrected it now!

ujjwal · Answer

@estudier

anonymous · Answer

Use cos^2 + sin^2 = 1 to rewrite, solve quadratic..

anonymous · Answer

change the $\large cos^2\theta \rightarrow 1-sin^2\theta $.
now you have a trignometric equation in quadratic form..
solve the quadratic.

ujjwal · Answer

I am being unable to solve the quadratic equation..

anonymous · Answer

You cannot solve x^2-x-2 where x= sin theta?

ujjwal · Answer

yeah, i can.. I was making a silly mistake...

anonymous · Answer

Happens to us all.....