How do you solve $2sin^2(x)+3cos(x)=0$ on the interval [0,2pi]?

Question

How do you solve $2sin^2(x)+3cos(x)=0$ on the interval [0,2pi]?

1 Answer

Impact of this question

36623 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-20T00:07:45

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

Explanation:

Replace in the equation $sin^2 x$ by $(1 - cos^2 x)$ -->
$2(1 - cos^2 x) + 3cos x = 0$
$-2cos^2 x + 3cos x + 2 = 0$
Solve this equation for cos x.
$D = d^2 = b^2 - 4ac = 9 + 16 = 25$ --> $d = +- 5$
$cos x = -b/(2a) +- d/(2a) = 3/4 +- 5/-4 = 3/4 +- 5/4$
a. $cos x = 8/4 = 2$ (rejected because > 1)
b. $cos x = -2/4 = -1/2$
$cos x = -1/2$ --> $x = +- (2pi)/3$
Answers for the interval $(0, 2pi)$ :
$(2pi)/3, (4pi)/3$ (co-terminal to $(-2pi)/3)$

Science

Math

Humanities

... and beyond

How do you solve $2sin^2(x)+3cos(x)=0$ on the interval [0,2pi]?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2sin^2(x)+3cos(x)=02sin^2(x)+3cos(x)=0 on the interval [0,2pi]?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $2sin^2(x)+3cos(x)=0$ on the interval [0,2pi]?