Question

How do I solve $2{\sin ^2}x - 1 = 0$?

Answer 1

Hint: The given equation is a sinusoidal equation; in this equation we have to find the value of x, which is the angle of sin function. The above equation is solved by using basic mathematical operations of division, square root and addition of the terms.
Using the basic mathematical operations we will solve the given equations.

Complete step-by-step solution:
Let us discuss the basic mathematical operations first and then we will proceed for the problem.
When an integer is shifted from RHS to LHS and LHS to RHS its sign changes, that is from positive to negative and negative to positive. When an equation contains a variable then we must separate the variable on one side and constant on one side. If the coefficient of variable is one then the value of the variable can be calculated directly, if the coefficient of variable is more than one, then that constant value must be divided with the other constant value which was separated.
Sine function is a sinusoidal function whose value keeps on varying between -1 and +1. According to the right triangle sin function with some angle x is equal to the perpendicular upon base.
Let’s come to the calculation part;
$ \Rightarrow 2{\sin ^2}x - 1 = 0$(We will shift 1 to RHS)
$ \Rightarrow 2{\sin ^2}x = 1$
$ \Rightarrow {\sin ^2}x = \dfrac{1}{2}$ (2 is shifted to divide the term on RHS)
$ \Rightarrow \sin x = \dfrac{1}{{\sqrt 2 }}$ (We have removed the square)
$ \Rightarrow x = {\sin ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$ (Sin inverse one by square root 2 is equal to $45^\circ$)
$ \Rightarrow x = {45^\circ}$

Therefore the value of x is equal to $45^\circ$.

Note: Like the $\sin {45^\circ}$ is equal to $\dfrac{1}{{\sqrt 2 }}$ , in the similar manner $\sin {30^\circ}$ is equal to $\dfrac{1}{2}$, $\sin {60^\circ}$ is equal to $\dfrac{{\sqrt 3 }}{2}$, $\sin {90^\circ}$ is equal to 1. Cosine too have different values the way sin has, $\cos {30^\circ}$ is equal to $\dfrac{{\sqrt 3 }}{2}$, $\cos {60^\circ}$ is equal to $\dfrac{1}{2}$, $\cos {45^\circ}$ is equal to $\dfrac{1}{{\sqrt 2 }}$, $\cos {90^\circ}$ is equal to zero.