Question

2sinx+1 = 0 help me :v

Answer 1

Answer:

$\displaystyle \large{ x = \frac{7\pi}{6} + 2\pi k , \frac{11\pi}{6} + 2\pi k \: \: \: \: \: (k \in \Z)}$

Step-by-step explanation:

First, isolate sinx.

$\displaystyle \large{2 \sin x + 1 = 0}$

Subtract both sides by 1; after that, divide both sides by 2.

$\displaystyle \large{2 \sin x + 1 - 1 = 0 - 1} \\ \displaystyle \large{2 \sin x = - 1} \\ \displaystyle \large{ \frac{2 \sin x}{2} = \frac{ - 1}{2} } \\ \displaystyle \large{ \sin x = - \frac{1}{2} }$

Now recall a unit circle. Sine is only negative in Q3 and Q4.

1/2 is 30° or π/6 for sine.

Finding the Q3 and Q4 angles.

Q3

Using π + π/6 = 6π/6 + π/6 = 7π/6

Q4

Using 2π - π/6 = 12π/6 - π/6 = 11π/6

Now we know that in Q3, sinx = -1/2 is 7π/6 and in Q4, it's 11π/6.

Thus:-

$\displaystyle \large{ x = \frac{7\pi}{6} , \frac{11\pi}{6} }$

But since you did not specific the interval, assume that we are finding general solutions.

From Identity:

$\displaystyle \large{ \sin x = x + 2\pi k \: \: \: (k \in \Z)}$

Therefore since x = 7π/6 and x = 11π/6:

$\displaystyle \large{ x = \frac{7\pi}{6} + 2\pi k , \frac{11\pi}{6} + 2\pi k}$

where k is any integers.