# How can i solve this equation: 2sen^2(t)-cos(t)-1=0 ?

May 20, 2018

If you meant ${\sin}^{2} \left(t\right)$ then you can use ${\sin}^{2} \left(t\right) = 1 - {\cos}^{2} \left(t\right)$ and you will get (after simplifications) ${\cos}^{2} \left(t\right) + \frac{1}{2} \cos \left(t\right) - \frac{1}{2} = 0$

#### Explanation:

After applying the hint you will get $2 \left(1 - {\cos}^{2} \left(t\right)\right) - \cos \left(t\right) - 1 = 0$
This is
${\cos}^{2} \left(t\right) + \frac{1}{2} \cos \left(t\right) - \frac{1}{2} = 0$
substituting $z = \cos \left(t\right)$ you have to solve ${z}^{2} + \frac{1}{2} z - \frac{1}{2} = 0$

May 20, 2018

$t = \left(2 k + 1\right) \pi$
$t = \pm \frac{p}{3} + 2 k \pi$

#### Explanation:

$2 {\sin}^{2} t - \cos t - 1 = 0$
Replace ${\sin}^{2} t$ by $\left(1 - {\cos}^{2} t\right)$
$2 - 2 {\cos}^{2} t - \cos t - 1 = 0$
Solve this quadratic equation for cos t:
$- 2 {\cos}^{2} t - \cos t + 1 = 0$
Since a - b + c = 0, use shortcut. The 2 real roots are:
$\cos t = - 1$, and $\cos t = - \frac{c}{a} = \frac{1}{2}$
a. cos t = -1
Unit circle gives -->$t = \pi + 2 k \pi = \left(2 k + 1\right) \pi$
b. $\cos t = \frac{1}{2}$
Trig table and unit circle give -->
$t = \pm \frac{\pi}{3} + 2 k \pi$

Aug 10, 2018

An odd multiple of $\pi \mathmr{and} \left(6 k \pm 1\right) \frac{\pi}{3} , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

#### Explanation:

using ${\sin}^{2} t = 1 - {\cos}^{2} t$, te equation becomes a quadratic in

$\cos t$, giving

cos t = -1, 1/2 = cos pi, cos pi/3 and these give

$t = 2 k \pi \pm \pi \mathmr{and} 2 k \pi \pm \frac{\pi}{3} , k = 0 , 1 , 2 , 3 , \ldots$

= an odd multiple of pi, ( 6k +- 1)pi/3.

The list near 0 is

#t = .....- 7/3pi, - 5/3pi,- pi, - pi/3, pi/3, pi, 5/3pi, ...

See graph, for t-intercepts as solutions.
graph{y-2 (sin(x))^2 +cos (x) + 1=0[-10 10 -5 5]}