Question

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

Answer 1

If you meant `sin^2(t)` then you can use `sin^2(t)=1-cos^2(t)` and you will get (after simplifications) `cos^2(t)+1/2cos(t)-1/2=0`

Explanation:

After applying the hint you will get `2(1-cos^2(t))-cos(t)-1=0`
This is
`cos^2(t)+1/2cos(t)-1/2=0`
substituting `z=cos(t)` you have to solve `z^2+1/2z-1/2=0`

Answer 2

`t = (2k+1)pi`
`t = +- p/3 + 2kpi`

Explanation:

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

Answer 3

An odd multiple of `pi and ( 6k +- 1)pi/3, k = 0, +- 1, +-2, +-3, ...`

Explanation:

using `sin^2t = 1 - cos^2t`, te equation becomes a quadratic in

`cost`, giving

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

`t = 2kpi +- pi and 2kpi +- pi/3, k = 0, 1, 2, 3, ...`

#= 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]}