Answer

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**Hint:**

Here, we have to find the value of the variable. We can solve the equation by using the trigonometric identity. Trigonometric equation is an equation involving one or more trigonometric ratios of unknown angles.

**Formula Used:**

We will use the following formulae:

1) Exponential Rule: \[{\left( {{a^x}} \right)^{\dfrac{1}{y}}} = {a^{\dfrac{x}{y}}}\]; \[{x^0} = 1\]

2) Logarithmic Rule: \[\log {a^b} = b\log a\]; \[{\log _a}a = 1\]

3) Trigonometric Identity: \[\sin (2\pi + \theta ) = \sin \theta ;\]

**Complete step by step solution:**

We are given with an equation \[\sin \left( {x + \dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^\theta }}}} \right) = \cos \left( {x + \dfrac{{{{\left( { - 1} \right)}^{16}}}}{2} - \dfrac{{{{\log }_2}\left( {\sqrt 8 } \right)}}{3}} \right)\]

First, we are converting the complex term inside the braces into a more simpler form.

Now, for the first term \[\dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^0}}}\], we will get

Exponential Rule: \[{\left( {{a^x}} \right)^{\dfrac{1}{y}}} = {a^{\dfrac{x}{y}}}\]and \[{x^0} = 1\]

By using the exponential rule, we have

\[ \Rightarrow \dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^0}}} = \dfrac{{{\pi ^3} + 2{\pi ^3}}}{{2{\pi ^2}}} + {\pi ^1}\]

Adding the like terms, we will get

\[ \Rightarrow \dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^0}}} = \dfrac{{3{\pi ^3}}}{{2{\pi ^2}}} + \pi \]

Dividing the term, we will get

\[ \Rightarrow \dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^0}}} = \dfrac{{3\pi }}{2} + \pi \]

By taking L.C.M., we will get

\[ \Rightarrow \dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^0}}} = \dfrac{{3\pi }}{2} + \pi \times \dfrac{2}{2}\]

\[ \Rightarrow \dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^0}}} = \dfrac{{3\pi }}{2} + \dfrac{{2\pi }}{2}\]

Adding the like terms, we get

\[ \Rightarrow \dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^0}}} = \dfrac{{5\pi }}{2}\]

Rewriting the equation, we get

\[ \Rightarrow \dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^0}}} = 2\pi + \dfrac{\pi }{2}\]

Now,

By Rewriting the equation, we get

\[ \Rightarrow \dfrac{{{{( - 1)}^{16}}}}{2} - \dfrac{{{{\log }_2}\sqrt 8 }}{3} = \dfrac{1}{2} - \dfrac{{{{\log }_2}\sqrt {{2^3}} }}{3}\]

\[ \Rightarrow \dfrac{{{{( - 1)}^{16}}}}{2} - \dfrac{{{{\log }_2}\sqrt 8 }}{3} = \dfrac{1}{2} - \dfrac{{{{\log }_2}{2^{\dfrac{3}{2}}}}}{3}\]

Logarithmic Rule: \[\log {a^b} = b\log a\]

By using logarithmic rule, we get

\[ \Rightarrow \dfrac{{{{( - 1)}^{16}}}}{2} - \dfrac{{{{\log }_2}\sqrt 8 }}{3} = \dfrac{1}{2} - \dfrac{{\dfrac{3}{2}{{\log }_2}2}}{3}\]

\[ \Rightarrow \dfrac{{{{( - 1)}^{16}}}}{2} - \dfrac{{{{\log }_2}\sqrt 8 }}{3} = \dfrac{1}{2} - \dfrac{{3{{\log }_2}2}}{{2 \cdot 3}}\]

Dividing both the terms, we get

\[ \Rightarrow \dfrac{{{{( - 1)}^{16}}}}{2} - \dfrac{{{{\log }_2}\sqrt 8 }}{3} = \dfrac{1}{2} - \dfrac{{{{\log }_2}2}}{2}\]

Logarithmic Rule: \[{\log _a}a = 1\]

By using logarithmic rule, we get

\[ \Rightarrow \dfrac{{{{( - 1)}^{16}}}}{2} - \dfrac{{{{\log }_2}\sqrt 8 }}{3} = \dfrac{1}{2} - \dfrac{1}{2}\]

\[ \Rightarrow \dfrac{{{{( - 1)}^{16}}}}{2} - \dfrac{{{{\log }_2}\sqrt 8 }}{3} = 0\]

Now, by using the equation and substituting the values, we get

\[ \Rightarrow \sin \left( {x + \dfrac{{{\pi ^3} + 2\sqrt {{\pi ^6}} }}{{{\pi ^2} + {\pi ^2}}} + {\pi ^{{\pi ^\theta }}}} \right) = \cos \left( {x + \dfrac{{{{\left( { - 1} \right)}^{16}}}}{2} - \dfrac{{{{\log }_2}\left( {\sqrt 8 } \right)}}{3}} \right)\]

\[ \Rightarrow \sin \left( {x + 2\pi + \dfrac{\pi }{2}} \right) = \cos \left( {x + 0} \right)\]

\[ \Rightarrow \sin \left( {2\pi + \dfrac{\pi }{2} + x} \right) = \cos \left( x \right)\]

Trigonometric Identity: \[\sin (2\pi + \theta ) = \sin \theta \]

Now, by using the trigonometric identity, we get

\[ \Rightarrow \sin \left( {\dfrac{\pi }{2} + x} \right) = \cos \left( x \right)\]

This equation holds true for all \[x \in {\bf{R}}\].

So, \[x = 0,\dfrac{\pi }{2}\].

**Therefore, the two values of \[x\] that satisfy the equation are \[0,\dfrac{\pi }{2}\]**

**Note:**

We should make sure of using the trigonometric identity, exponential rule, the logarithmic rule at the right place. A trigonometric equation will also have a general solution expressing all the values which would satisfy the given equation, and it is expressed in a generalized form in terms of ‘n’. Thus the trigonometric equation always possess various solutions.

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