How do you solve the equation on the interval [0,2pi) for $2(cos(t))^2-cos(t)-1=0$ ?

Question

How do you solve the equation on the interval [0,2pi) for $2(cos(t))^2-cos(t)-1=0$ ?

1 Answer

Impact of this question

16564 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-14T16:26:22

$0, (2pi)/3, (4pi)/3, 2pi$

Explanation:

Solve: $2cos^2 t - cos t - 1 = 0$ for cos t.
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 --> t = 0 or t = $2pi$
b. $cos t = - 1/2$
Trig table and unit circle -->
$cos t = - 1/2$ --> $t = +- (2pi)/3$
Arc $(4pi)/3$ is co-terminal to arc $(-2pi/3)$ .
Answer for $(0, 2pi):$
$0, (2pi)/3, (4pi)/3, 2pi$

Science

Math

Humanities

... and beyond

How do you solve the equation on the interval [0,2pi) for $2(cos(t))^2-cos(t)-1=0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve the equation on the interval [0,2pi) for 2(cos(t))^2-cos(t)-1=0 2(cos(t))^2-cos(t)-1=0 ?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve the equation on the interval [0,2pi) for $2(cos(t))^2-cos(t)-1=0$ ?