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Chapter 5: Problem 40

Solve the multiple-angle equation. $$2 \sin 2 x+\sqrt{3}=0$$

Short Answer

Expert verified
The solutions for the given equation are \(x = 2蟺/3 + 蟺k\) and \(x = 5蟺/6 + 蟺k\) for all integers \(k\).

Step by step solution

01

Simplify the Equation

The given equation is \(2 \sin 2x + \sqrt{3} = 0\). Rearrange this equation to give \( \sin 2x = -\sqrt{3}/2\). This is achieved by moving \(\sqrt{3}\) to the other side and dividing by 2.
02

Implement Trigonometric Principles

Now, we know that \(\sin(胃) = -\sqrt{3}/2\) at \(胃 = 4蟺/3\) and \(胃 = 5蟺/3\). Therefore, we can say that \(2x = 4蟺/3 + 2蟺k\) or \(2x = 5蟺/3 + 2蟺k\), for all integers \(k\), because of the 2蟺 periodicity of the sine function.
03

Solve for x

By dividing each equation by 2, we have \(x = 2蟺/3 + 蟺k\) or \(x = 5蟺/6 + 蟺k\) for all integers \(k\).

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