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Question: How do you solve \(2{\sin ^2}x - 1 = 0\)?...

How do you solve 2sin⁡2x−1=02{\sin ^2}x - 1 = 0?

Explanation

Solution

In order to determine the value of the above question, first simplify the expression by taking constants on right hand side and taking inverse of sine on both sides of the equation to pull out x from inside of the sine. Now we simply have to find the value of the inverse of sine of ±12 \pm \dfrac{1}{{\sqrt 2 }}. The sine function is positive in both 1st and 2nd quadrant and negative in 3rd and 4th quadrant.

Complete step by step answer:
We are given a trigonometric expression 2sin⁡2x−1=02{\sin ^2}x - 1 = 0
First we will simplify the expression by taking constants on right hand side
2sin⁡2x=−12{\sin ^2}x = - 1
Now, dividing both sides of the equation with number 2
⇒22sin⁡2x=12 ⇒sin⁡2x=12 ⇒sin⁡x=±12  \Rightarrow \dfrac{2}{2}{\sin ^2}x = \dfrac{1}{2} \\\ \Rightarrow {\sin ^2}x = \dfrac{1}{2} \\\ \Rightarrow \sin x = \pm \dfrac{1}{{\sqrt 2 }} \\\
Taking the inverse of sine on both sides of the equation to pull out variable from inside the sine,we get
⇒sin⁡−1(sin⁡x)=sin⁡−1(±12) ⇒x=sin⁡−1(±12) ⇒x=sin⁡−1(12) ⇒x=sin⁡−1(−12)  \Rightarrow {\sin ^{ - 1}}(\sin x) = {\sin ^{ - 1}}\left( { \pm \dfrac{1}{{\sqrt 2 }}} \right) \\\ \Rightarrow x = {\sin ^{ - 1}}\left( { \pm \dfrac{1}{{\sqrt 2 }}} \right) \\\ \Rightarrow x = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\\ \Rightarrow x = {\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) \\\
The value of sin⁡−1(12){\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) at π4\dfrac{\pi }{4} and 3π4\dfrac{{3\pi }}{4}.
Since, we know that the sine function is always positive in the both first and third quadrants.
Similarly, The value of sin⁡−1(−12){\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right) at 5π4\dfrac{{5\pi }}{4} and 7π4\dfrac{{7\pi }}{4}.
For the general case, x=π4+kπ2x = \dfrac{\pi }{4} + k\dfrac{\pi }{2} k is an integer

Therefore the solution to 2sin⁡2x−1=02{\sin ^2}x - 1 = 0 is at x=π4,3π4,5π4,7π4x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}in the range(0,2π)(0,2\pi ) or in general x=π4+kπ2x = \dfrac{\pi }{4} + k\dfrac{\pi }{2} for all integer values of k.

Additional information:
1. In Mathematics the inverse trigonometric functions (every so often additionally called anti-trigonometric functions or cyclomatic function) are the reverse elements of the mathematical functions In particular, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are utilized to get a point from any of the point's mathematical proportions. Reverse trigonometric functions are generally utilized in designing, route, material science, and calculation.
2. In inverse trigonometric function, the domain are the ranges of corresponding trigonometric functions and the range are the domain of the corresponding trigonometric function.

Note: Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
Periodic Function = A function f(x)f(x) is said to be a periodic function if there exists a real number T > 0 such that f(x+T)=f(x)f(x + T) = f(x) for all x.