If ${{\sin }^{2}}\left( 3x+45 \right)+{{\cos }^{2}}\left( 2x+60 \right)=1$, then x equals
[a] 60
[b] 30
[c] 15
[d] 0

Hint: Subtract ${{\cos }^{2}}\left( 2x+60 \right)$ from both sides of the equation. Use the fact that ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. Use the fact that if ${{a}^{2}}={{b}^{2}},$ then $a=\pm b$. Use the fact that if $\sin x=\sin y$, then $x=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}$. Hence form an equation in x. Solve for x and hence find which of the given options are correct.

Complete step-by-step answer:
We have \[{{\sin }^{2}}\left( 3x+45 \right)+{{\cos }^{2}}\left( 2x+60 \right)=1\]
Subtracting ${{\cos }^{2}}\left( 2x+60 \right)$ from both sides, we get
${{\sin }^{2}}\left( 3x+45 \right)=1-{{\cos }^{2}}\left( 2x+60 \right)$
We know that ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $
Using the above identity, we get
${{\sin }^{2}}\left( 3x+45 \right)={{\sin }^{2}}\left( 2x+60 \right)$
We know that if ${{a}^{2}}={{b}^{2}},$ then $a=\pm b$
Hence, we have
$\sin \left( 3x+45 \right)=\pm \sin \left( 2x+60 \right)$
Taking the positive sign, we get
$\sin \left( 3x+45 \right)=\sin \left( 2x+60 \right)$
We know that if $\sin x=\sin y$, then $x=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}$.
Hence, we have
$3x+45=n\pi +{{\left( -1 \right)}^{n}}\left( 2x+60 \right)$
Since there are no $\pi $ terms in the options, taking n=0, we get
$3x+45=2x+60$
Subtracting 45 from both sides of the equation, we get
$3x=2x+15$
Subtracting 2x from both sides of the equation, we get
$x=15$
Hence option [c] is correct.
Taking the negative sign, we get
$\sin \left( 3x+45 \right)=-\sin \left( 2x+60 \right)$
We know that $\sin \left( -x \right)=-\sin x$
Hence, we have
$\sin \left( 3x+45 \right)=\sin \left( -2x-60 \right)$
We know that if $\sin x=\sin y$, then $x=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}$.
Hence, we have
$3x+45=n\pi +{{\left( -1 \right)}^{n}}\left( -2x-60 \right),n\in \mathbb{Z}$
Since there are no $\pi $ terms in options, taking n = 0, we get
$3x+45=-2x-60$
Adding 2x on both sides, we get
$5x+45=-60$
Subtracting -45 from both sides, we get
$5x=-105$
Dividing by 5 on both sides of the equation, we get
x =- 21
Hence option [c] is the only correct answer.

Note: We can also check option wise which of the options is correct.
Option [a]: 60
We have $\sin \left( 3x+45 \right)=\sin \left( 180+45 \right)=\sin 225$ and $\cos \left( 2x+60 \right)=\cos \left( 120+60 \right)=\cos 180$
Since ${{\sin }^{2}}225+{{\cos }^{2}}180\ne 1$, option [a] is not correct.
Option [b]: 30
We have $\sin \left( 3x+45 \right)=\sin \left( 90+45 \right)=\sin 135$ and $\cos \left( 2x+60 \right)=\cos \left( 60+60 \right)=\cos 120$
Since ${{\sin }^{2}}135+{{\cos }^{2}}120\ne 1$, option [b] is incorrect.
Option [c]: 15
We have
$\sin \left( 3x+45 \right)=\sin \left( 45+45 \right)=\sin 90$ and $\cos \left( 2x+60 \right)=\cos \left( 30+60 \right)=\cos 90$
Since ${{\sin }^{2}}90+{{\cos }^{2}}90=1$, option [c] is correct
Option [d]: 0
We have $\sin \left( 3x+45 \right)=\sin 45$ and $\cos \left( 2x+60 \right)=\cos 60$
Since ${{\sin }^{2}}45+{{\cos }^{2}}60\ne 1$, option [d] is incorrect.
Hence option [c] is the only correct answer.