Solve sin^2 x-5 sinx+6=0 for all solutions [0,2pi)?

Question

Solve sin^2 x-5 sinx+6=0 for all solutions [0,2pi)?

1 Answer

Impact of this question

8591 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-17T06:28:45

No real solutions.

Explanation:

$sin^2(x)-5sin(x)+6 = 0$

$=> (sin(x)-2)(sin(x)-3) = 0$

$=> sin(x)-2 = 0 or sin(x)-3 = 0$

$=> sin(x) = 2 or sin(x) = 3$

However, $sin(x) in [-1, 1]$ for all $x in RR$ , meaning neither of the above equations has a solution in $[0, 2pi)$ . Indeed, the given equation has no real solutions.

Science

Math

Humanities

... and beyond

Solve sin^2 x-5 sinx+6=0 for all solutions [0,2pi)?

1 Answer

Explanation:

Related questions

Impact of this question