How do you solve $2 {sin}^{2} x = sin x$ $2sin^2x=Sinx$ over the interval 0 to 2pi?

Question

How do you solve $2 {sin}^{2} x = sin x$ $2sin^2x=Sinx$ over the interval 0 to 2pi?

1 Answer

Impact of this question

30340 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-17T07:08:06

0, $π$ $pi$ , $2 π$ $2pi$ , $\frac{π}{6}$ $pi/6$ and $5 \frac{π}{6}$ $5pi/6$ ,

Explanation:

Rearranging, sinx (2 sinx - 1) = 0.
In [0, 2 $π$ $pi$ ],sin x = 0 for x = 0, $π$ $pi$ and $2 π$ $2pi$ .
The other factor = 0 gives the other two solutions.
sin $5 \frac{π}{6}$ $5pi/6$ = sin ( $π - \frac{π}{6}$ $pi - pi/6$ ) = sin $\frac{π}{6}$ $pi/6$ = 1/2.

Science

Math

Humanities

... and beyond

How do you solve $2 {sin}^{2} x = sin x$ $2sin^2x=Sinx$ over the interval 0 to 2pi?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2sin^2x=Sinx2sin2x=sinx2sin^2x=Sinx over the interval 0 to 2pi?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $2 {sin}^{2} x = sin x$ $2sin^2x=Sinx$ over the interval 0 to 2pi?