x-0-2pi-2cos-2-x-sinx-1-0-x-

Question Number 11249 by uni last updated on 18/Mar/17

x \in (0, 2 π)

{2 \cos}^{2} x + s i n x - 1 = 0 \Rightarrow Σ x = ?

Answered by ajfour last updated on 18/Mar/17

2 - 2 \sin^{2} x + \sin x - 1 = 0

2 \sin^{2} x - \sin x - 1 = 0

(2 \sin x + 1) (\sin x - 1) = 0

\sin x = - \frac{1}{2}, 1

as x \in (0, 2 π)

x = \frac{π}{2}, π + \frac{π}{6}, 2 π - \frac{π}{6}

Σ x = \frac{7 π}{2} .

Answered by prakash jain last updated on 18/Mar/17

2 (1 - \sin^{2} x) + \sin x - 1 = 0

2 - {2 \sin}^{2} x + \sin x - 1 = 0

{2 \sin}^{2} x - \sin x - 1 = 0

{2 \sin}^{2} x - 2 \sin x + \sin x - 1 = 0

2 \sin x (\sin x - 1) + (\sin x - 1) = 0

(\sin x - 1) (2 \sin x + 1) = 0

\sin x = 1 \Rightarrow x = \frac{π}{2}

\sin x = - \frac{1}{2} \Rightarrow x = \frac{5 π}{6}, \frac{11 π}{6}

Σ x = \frac{π}{2} + \frac{5 π}{6} + \frac{11 π}{6} = \frac{7 π}{2}