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Question: If \({{\sin }^{2}}\left( 3x+45 \right)+{{\cos }^{2}}\left( 2x+60 \right)=1\), then x equals [a] 60...

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

Explanation

Solution

Hint: Subtract cos⁡2(2x+60){{\cos }^{2}}\left( 2x+60 \right) from both sides of the equation. Use the fact that sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta . Use the fact that if a2=b2,{{a}^{2}}={{b}^{2}}, then a=±ba=\pm b. Use the fact that if sin⁡x=sin⁡y\sin x=\sin y, then x=nπ+(−1)ny,n∈Zx=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(3x+45)+cos⁡2(2x+60)=1{{\sin }^{2}}\left( 3x+45 \right)+{{\cos }^{2}}\left( 2x+60 \right)=1
Subtracting cos⁡2(2x+60){{\cos }^{2}}\left( 2x+60 \right) from both sides, we get
sin⁡2(3x+45)=1−cos⁡2(2x+60){{\sin }^{2}}\left( 3x+45 \right)=1-{{\cos }^{2}}\left( 2x+60 \right)
We know that cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta
Using the above identity, we get
sin⁡2(3x+45)=sin⁡2(2x+60){{\sin }^{2}}\left( 3x+45 \right)={{\sin }^{2}}\left( 2x+60 \right)
We know that if a2=b2,{{a}^{2}}={{b}^{2}}, then a=±ba=\pm b
Hence, we have
sin⁡(3x+45)=±sin⁡(2x+60)\sin \left( 3x+45 \right)=\pm \sin \left( 2x+60 \right)
Taking the positive sign, we get
sin⁡(3x+45)=sin⁡(2x+60)\sin \left( 3x+45 \right)=\sin \left( 2x+60 \right)
We know that if sin⁡x=sin⁡y\sin x=\sin y, then x=nπ+(−1)ny,n∈Zx=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}.
Hence, we have
3x+45=nπ+(−1)n(2x+60)3x+45=n\pi +{{\left( -1 \right)}^{n}}\left( 2x+60 \right)
Since there are no π\pi terms in the options, taking n=0, we get
3x+45=2x+603x+45=2x+60
Subtracting 45 from both sides of the equation, we get
3x=2x+153x=2x+15
Subtracting 2x from both sides of the equation, we get
x=15x=15
Hence option [c] is correct.
Taking the negative sign, we get
sin⁡(3x+45)=−sin⁡(2x+60)\sin \left( 3x+45 \right)=-\sin \left( 2x+60 \right)
We know that sin⁡(−x)=−sin⁡x\sin \left( -x \right)=-\sin x
Hence, we have
sin⁡(3x+45)=sin⁡(−2x−60)\sin \left( 3x+45 \right)=\sin \left( -2x-60 \right)
We know that if sin⁡x=sin⁡y\sin x=\sin y, then x=nπ+(−1)ny,n∈Zx=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}.
Hence, we have
3x+45=nπ+(−1)n(−2x−60),n∈Z3x+45=n\pi +{{\left( -1 \right)}^{n}}\left( -2x-60 \right),n\in \mathbb{Z}
Since there are no π\pi terms in options, taking n = 0, we get
3x+45=−2x−603x+45=-2x-60
Adding 2x on both sides, we get
5x+45=−605x+45=-60
Subtracting -45 from both sides, we get
5x=−1055x=-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⁡(3x+45)=sin⁡(180+45)=sin⁡225\sin \left( 3x+45 \right)=\sin \left( 180+45 \right)=\sin 225 and cos⁡(2x+60)=cos⁡(120+60)=cos⁡180\cos \left( 2x+60 \right)=\cos \left( 120+60 \right)=\cos 180
Since sin⁡2225+cos⁡2180≠1{{\sin }^{2}}225+{{\cos }^{2}}180\ne 1, option [a] is not correct.
Option [b]: 30
We have sin⁡(3x+45)=sin⁡(90+45)=sin⁡135\sin \left( 3x+45 \right)=\sin \left( 90+45 \right)=\sin 135 and cos⁡(2x+60)=cos⁡(60+60)=cos⁡120\cos \left( 2x+60 \right)=\cos \left( 60+60 \right)=\cos 120
Since sin⁡2135+cos⁡2120≠1{{\sin }^{2}}135+{{\cos }^{2}}120\ne 1, option [b] is incorrect.
Option [c]: 15
We have
sin⁡(3x+45)=sin⁡(45+45)=sin⁡90\sin \left( 3x+45 \right)=\sin \left( 45+45 \right)=\sin 90 and cos⁡(2x+60)=cos⁡(30+60)=cos⁡90\cos \left( 2x+60 \right)=\cos \left( 30+60 \right)=\cos 90
Since sin⁡290+cos⁡290=1{{\sin }^{2}}90+{{\cos }^{2}}90=1, option [c] is correct
Option [d]: 0
We have sin⁡(3x+45)=sin⁡45\sin \left( 3x+45 \right)=\sin 45 and cos⁡(2x+60)=cos⁡60\cos \left( 2x+60 \right)=\cos 60
Since sin⁡245+cos⁡260≠1{{\sin }^{2}}45+{{\cos }^{2}}60\ne 1, option [d] is incorrect.
Hence option [c] is the only correct answer.