Question

How do you solve the equation $ \sqrt{2}\sin x-1=0 $ for $ -{{360}^{\circ }}\text{ }<\text{ }x\le {{360}^{\circ }} $ ?

Answer 1

Hint: In this question, we need to solve the equation $ \sqrt{2}\sin x-1=0 $ . For all values of x such that x lies between $ -{{360}^{\circ }}\text{ and }{{360}^{\circ }} $ . In radian form, x must lie between $ -2\pi \text{ and 2}\pi $ . We will first arrange the equation and find value of sinx. Then we will take the known value of x for which equation satisfies using the trigonometric ratio table. After that, we will use the general solution of sinx by the formula that if $ \sin x=\sin y\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}y $ . Using this we will find value of n and find value of x lying between $ -2\pi \text{ and 2}\pi $ .

Complete step by step answer:
Here we are given the equation as $ \sqrt{2}\sin x-1=0 $ . We need to find the value of x lying in the interval of $ -{{360}^{\circ }}\text{ to }{{360}^{\circ }} $ . Let us change interval into radian. We know $ {{360}^{\circ }} $ is $ 2\pi $ . So we have interval as $ \left( -\pi ,\pi \right) $ .
Now our equation is $ \sqrt{2}\sin x-1=0 $ .
Taking 1 to the other side we get $ \sqrt{2}\sin x=1 $ .
Dividing both sides by $ \sqrt{2} $ we get $ \sin x=\dfrac{1}{\sqrt{2}} $ .
Now let us look at the trigonometric ratio table to find one of the value of x such that $ \sin x=\dfrac{1}{\sqrt{2}} $ .

$ \theta \to $	$ {{0}^{\circ }} $	$ \dfrac{\pi }{6} $	$ \dfrac{\pi }{4} $	$ \dfrac{\pi }{3} $	$ \dfrac{\pi }{2} $
sin	0	$ \dfrac{1}{2} $	$ \dfrac{1}{\sqrt{2}} $	$ \dfrac{\sqrt{3}}{2} $	1

We see that $ \sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}} $ .
So we can use $ \sin x=\sin \dfrac{\pi }{4} $ .
One of the value of x will be $ \dfrac{\pi }{4} $ . But we need all the values lying between $ -2\pi \text{ and 2}\pi $ .
So let us find the general value of x.
We know that if sinx = siny then $ x=n\pi +{{\left( -1 \right)}^{n}}y,n\in z $ .
Therefore, if $ \sin x=\sin \dfrac{\pi }{4} $ , then $ x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{4},n\in z $ .
Now let us put values of n to find values of x.
Putting n = 0 we get $ x=\dfrac{\pi }{4} $ .
Putting n = 1 we get $ x=\pi -\dfrac{\pi }{4}=\dfrac{3\pi }{4} $ (Lies between $ \left( -2\pi ,2\pi \right) $ )
Putting n = 2 we get $ x=2\pi +\dfrac{\pi }{4}=\dfrac{9\pi }{4} $ greater than $ 2\pi $ .
Putting n = -1 we get $ x=-\pi -\dfrac{\pi }{4}=\dfrac{-5\pi }{4} $ (Lies between $ \left( -2\pi ,2\pi \right) $ )
Putting n = -2 we get $ x=-2\pi +\dfrac{\pi }{4}=\dfrac{-7\pi }{4} $ (Lies between $ \left( -2\pi ,2\pi \right) $ )
Rest all of the values will lie outside $ \left( -2\pi ,2\pi \right) $ so value of x are $ \dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{-5\pi }{4}\text{ and }\dfrac{-7\pi }{4} $ .
Let us convert them into degree to match the units with question.
Taking $ \pi $ as $ {{180}^{\circ }} $ we get,
 $ \begin{align}
  & \dfrac{\pi }{4}=\dfrac{{{180}^{\circ }}}{4}={{45}^{\circ }} \\
 & \dfrac{3\pi }{4}=3\times \dfrac{{{180}^{\circ }}}{4}={{135}^{\circ }} \\
 & \dfrac{-5\pi }{4}=-5\times \dfrac{{{180}^{\circ }}}{4}=-{{225}^{\circ }} \\
 & \dfrac{-7\pi }{4}=-7\times \dfrac{{{180}^{\circ }}}{4}=-{{315}^{\circ }} \\
\end{align} $
Therefore, the required value of x lying between $ -{{360}^{\circ }}\text{ }<\text{ }x\le {{360}^{\circ }} $ are $ {{45}^{\circ }},{{135}^{\circ }},-{{225}^{\circ }},-{{315}^{\circ }} $ .
Note:
Students should keep in mind the trigonometric ratio table to solve the sum. Take care of signs while finding different values of x. As the question has angles in degrees so our answer should also be in degree. Keep in mind the general solutions of all trigonometric functions.