Solve the following equation for all radian solutions and if 0≤x≤2π. Give all answers as exact values in radians. 0≤x≤2π x=7π6,π2,11π6 rad

Given equation: 2sin2x−sin⁡x−1=0 2sin2x−2sin⁡x+sin⁡x−1=0 2sin⁡x(sin⁡x−1)+1(sin⁡x−1)=0 (sin⁡x−1)(2sin⁡x+1)=0 sin⁡x−1=0 or 2sin⁡x+1=0 sin⁡x=1 or 2sin⁡x=−1 ⇒sin=−12 sin⁡x=1 x=2π+π2, where is any integer sin⁡x is negative in quadrant 3 and 4. Then general solution is sin⁡x=−12 ⇒x=7π6+2kπ,11π6+2kπ for any inter In 0≤x&lt;2π, put x=0 we get x=π2,7π6,11π6

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sanuluy

Answered question

2021-09-12

Solve the following equation for all radian solutions and if

0 \leq x \leq 2 π

. Give all answers as exact values in radians.

0 \leq x \leq 2 π

x = \frac{7 π}{6}, \frac{π}{2}, \frac{11 π}{6} r a d

Answer & Explanation

Tasneem Almond

Skilled2021-09-13Added 91 answers

Given equation:
$2 \sin^{2} x - \sin x - 1 = 0$
$2 \sin^{2} x - 2 \sin x + \sin x - 1 = 0$
$2 \sin x (\sin x - 1) + 1 (\sin x - 1) = 0$
$(\sin x - 1) (2 \sin x + 1) = 0$
$\sin x - 1 = 0$ or $2 \sin x + 1 = 0$
$\sin x = 1$ or $2 \sin x = - 1$
$\Rightarrow \sin = \frac{- 1}{2}$
$\sin x = 1$
$x = 2 π + \frac{π}{2}$ , where is any integer
$\sin x$ is negative in quadrant 3 and 4.
Then general solution is
$\sin x = \frac{- 1}{2}$
$\Rightarrow x = \frac{7 π}{6} + 2 k π, \frac{11 π}{6} + 2 k π$ for any inter
In $0 \leq x < 2 π$ , put x=0 we get
$x = \frac{π}{2}, \frac{7 π}{6}, \frac{11 π}{6}$

Jeffrey Jordon

Expert2021-12-11Added 2605 answers

Answer is given below (on video)

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