2cos^2x-sinx=-1?

Question

2cos^2x-sinx=-1?

1 Answer

Impact of this question

3479 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-17T14:06:28

$\frac{π}{6}; \frac{5 π}{6}; \frac{3 π}{2}$ $pi/6; (5pi)/6; (3pi)/2$

Explanation:

Solve the equation:
$2 {cos}^{2} x - sin x + 1 = 0$ $2cos^2 x - sin x + 1 = 0$
Replace in the equation ${cos}^{2} x$ $cos^2 x$ by $(1 - {sin}^{2} x)$ $(1 - sin^2 x)$ -->
$2 - 2 {sin}^{2} x - sin x - 1 = 0$ $2 - 2sin^2 x - sin x - 1 = 0$
Solve this quadratic equation for sin x -->
$- 2 {sin}^{2} x - sin x + 1 = 0$ $-2sin^2 x - sin x + 1 = 0$
Since a - b + c = 0, use shortcut. The 2 real roots are:
sin x = -1 and $sin x = - \frac{c}{a} = \frac{1}{2}$ $sin x = - c/a = 1/2$
a. sin x = - 1
Unit circle gives --> $x = \frac{3 π}{2} + 2 k π$ $x = (3pi)/2 + 2kpi$
b. $sin x = \frac{1}{2}$ $sin x = 1/2$
Trig table and unit circle give 2 solutions for x -->
$x = \frac{π}{6}$ $x = pi/6$ , and $x = \frac{5 π}{6}$ $x = (5pi)/6$

Science

Math

Humanities

... and beyond

2cos^2x-sinx=-1?

1 Answer

Explanation:

Related questions

Impact of this question