Question

How do u solve this equation: \[{\csc ^4}\left( {2u} \right) - 4 = 0\]?

Answer 1

Hint: Trigonometric functions are simply defined as the functions of an angle of a triangle, it means that the relationship between the angles and sides of a triangle are given by these trigonometric functions. To solve the given trigonometric equation, factor out the terms, which is of the form; \[{a^2} - {b^2}\], next apply trigonometric identity formulas for cosec, as we know that sec, csc and cot are derived from primary functions of sin, cos and tan, hence using these functions we need to solve the given equation.

Formula used:
\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]

Complete step by step solution:
Given,
\[{\csc ^4}\left( {2u} \right) - 4 = 0\] …………………….. 1
Factor as:\[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\] ……………………. 2
 where; \[a = {\csc ^4}\left( {2u} \right)\] and \[b = 4\]
Now, substitute the value of a and b in terms of equation 2 as:
\[ \Rightarrow \left( {{{\csc }^4}\left( {2u} \right) - 2} \right)\left( {{{\csc }^4}\left( {2u} \right) + 2} \right) = 0\]
To find the values where the above equation is true, we must set both factors equal to 0:
\[{\csc ^2}\left( {2u} \right) - 2 = 0\] and \[{\csc ^2}\left( {2u} \right) + 2 = 0\] ……………… 3
Substitute, \[{\csc ^2}\left( {2u} \right) = \dfrac{1}{{{{\sin }^2}\left( {2u} \right)}}\] in equation 3 we get:
\[ \Rightarrow \dfrac{1}{{{{\sin }^2}\left( {2u} \right)}} - 2 = 0\] and \[\dfrac{1}{{{{\sin }^2}\left( {2u} \right)}} + 2 = 0\] ……………... 4
Add two to both sides of the first equation and subtract 2 from both sides of the second equation in equation 4 as:
\[ \Rightarrow \dfrac{1}{{{{\sin }^2}\left( {2u} \right)}} = 2\] and \[\dfrac{1}{{{{\sin }^2}\left( {2u} \right)}} = - 2\] ……………….. 5
Invert both sides of both equations in equation 5 we get:
\[ \Rightarrow {\sin ^2}\left( {2u} \right) = \dfrac{1}{2}\] and \[{\sin ^2}\left( {2u} \right) = - \dfrac{1}{2}\]
Take the square root of both sides of both equations:
\[ \Rightarrow \sin \left( {2u} \right) = \pm \dfrac{{\sqrt 2 }}{2}\] and \[\sin \left( {2u} \right) = \pm \dfrac{{\sqrt 2 }}{2}i\]
Take the inverse sine of both sides:
\[ \Rightarrow 2u = \pm {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{2}} \right)\] and \[2u = \pm {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{2}i} \right)\]
The first equation is well known i.e., \[{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{2}} \right) = \dfrac{\pi }{4}\] and the second equation becomes the inverse hyperbolic sine:
 \[ \Rightarrow 2u = \pm \dfrac{\pi }{4}\] and \[2u = \pm i{\sinh ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{2}} \right)\]
It will be periodic at integer multiples of π as:
\[ \Rightarrow 2u = n\pi \pm \dfrac{\pi }{4}\] and \[2u = n\pi \pm i{\sinh ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{2}} \right)\]
Divide by 2 we get:
 \[ \Rightarrow u = n\dfrac{\pi }{2} \pm \dfrac{\pi }{8}\] and \[u = n\dfrac{\pi }{2} \pm i\dfrac{1}{2}{\sinh ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{2}} \right)\]

Note: We must know that, the basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant and the key point to solve the values of any trigonometric function is to note all formulas of trigonometric identities functions related to the given equation and then substitute the formulas in the equation to find the terms asked.