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Chapter 1: Problem 41

Solve the following equations. $$\sqrt{2} \sin x-1=0$$

Short Answer

Expert verified
Answer: The general solutions for the equation are $$x_1 = \frac{\pi}{4} + 2n\pi$$ and $$x_2 = \frac{3\pi}{4} + 2n\pi$$, where n is an integer.

Step by step solution

01

Isolate sin(x) function

To isolate the sin(x) function, we need to get rid of the square root of 2 and the 1 on the left-hand side of the equation. To do this, add 1 to both sides of the equation and then divide by the square root of 2: $$\sqrt{2} \sin x - 1 + 1 = 0 + 1$$ $$\sqrt{2} \sin x = 1$$ Now, divide both sides by the square root of 2: $$\sin x = \frac{1}{\sqrt{2}}$$
02

Find the inverse sin values

To find the x values, we now take the inverse sin (sin^(-1)) of both sides: $$x = \arcsin{ \left(\frac{1}{\sqrt{2}} \right)}$$ The inverse sin function (arcsin) yields two possible values: - Principal value: $$x_1 = \frac{\pi}{4}$$ - Co-principal value: $$x_2 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$ These x values are solutions within the first trigonometric period [0, 2蟺).
03

Find the general solution

The general solution for a sine function can be written as: $$x = x_0 + 2n\pi$$ Where x_0 is the principal value, and n is an integer. For our principal value, the general solution is: $$x_1 = \frac{\pi}{4} + 2n\pi$$ For our co-principal value, the general solution is: $$x_2 = \frac{3\pi}{4} + 2n\pi$$ These two general solutions represent all solutions for the original equation $$\sqrt{2} \sin x - 1 = 0$$.

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