Answer :
Subtract sqrt 3 to isolate 2sin x
Divide by 2 to isolate sin x
Take the arcsin of each side to isolate x
Thinking of a 30-60-90 triangle, we know that x sqrt 3 is opposite 60° and 2x acts as the hypotenuse, thus it will be some variation of 60°.
In order to produce a negative result, x must be -60° (-π/3)
It is worth noting that this is a restricted value. Any addition/subtraction of 2π from this answer will produce the same result, thus your full answer is the above plus or minus 2kπ where k ∈ Z (k is an interger)
[tex]2\sin x+\sqrt{3}=0\\\\2\sin x=-\sqrt{3}\\\\\sin x=-\frac{\sqrt{3}}2\\\\\sin^{-1}(\frac{\sqrt{3}}2)=x\\\\\boxed{x=-\frac{\pi}3\pm2k\pi,\ k\in \mathbb{Z}}\ (-\frac{\pi}3=-60\°)[/tex]
Divide by 2 to isolate sin x
Take the arcsin of each side to isolate x
Thinking of a 30-60-90 triangle, we know that x sqrt 3 is opposite 60° and 2x acts as the hypotenuse, thus it will be some variation of 60°.
In order to produce a negative result, x must be -60° (-π/3)
It is worth noting that this is a restricted value. Any addition/subtraction of 2π from this answer will produce the same result, thus your full answer is the above plus or minus 2kπ where k ∈ Z (k is an interger)
[tex]2\sin x+\sqrt{3}=0\\\\2\sin x=-\sqrt{3}\\\\\sin x=-\frac{\sqrt{3}}2\\\\\sin^{-1}(\frac{\sqrt{3}}2)=x\\\\\boxed{x=-\frac{\pi}3\pm2k\pi,\ k\in \mathbb{Z}}\ (-\frac{\pi}3=-60\°)[/tex]
