The solution of the equation $\sin x + 3\sin 2x + \sin 3x = \cos x + 3\cos 2x + \cos 3x$ in the interval $0 \leqslant x \leqslant 2\pi $ are
$
(a){\text{ }}\dfrac{\pi }{3},\dfrac{{5\pi }}{8},\dfrac{{2\pi }}{3} \\
(b){\text{ }}\dfrac{\pi }{8},\dfrac{{5\pi }}{8},\dfrac{{9\pi }}{8},\dfrac{{13\pi }}{8} \\
(c){\text{ }}\dfrac{{4\pi }}{3},\dfrac{{9\pi }}{3},\dfrac{{2\pi }}{3},\dfrac{{13\pi }}{8} \\
(d){\text{ }}\dfrac{\pi }{8},\dfrac{{5\pi }}{8},\dfrac{{9\pi }}{3},\dfrac{{4\pi }}{3} \\
$
Answer
360.9k+ views
Hint – In this question we have to find the solution of the given equation, use trigonometric identities like $\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ to simplify for the value of x, but keep in mind that the solution is asked to be found only for the interval of \[0 \leqslant x \leqslant 2\pi \].
Complete step-by-step answer:
Given equation is
$\sin x + 3\sin 2x + \sin 3x = \cos x + 3\cos 2x + \cos 3x$
$ \Rightarrow \sin x + \sin 3x + 3\sin 2x = \cos x + \cos 3x + 3\cos 2x$
We have to find out the solution of this equation in the interval\[0 \leqslant x \leqslant 2\pi \].
Now as we know
$
\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) \\
\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) \\
$
So, use this property in the above equation we have,
$ \Rightarrow \sin x + \sin 3x + 3\sin 2x = \cos x + \cos 3x + 3\cos 2x$
$ \Rightarrow 2\sin \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right) + 3\sin 2x = 2\cos \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right) + 3\cos 2x$
Now simplify the above equation we have,
$ \Rightarrow 2\sin 2x\cos x + 3\sin 2x = 2\cos 2x\cos x + 3\cos 2x$ $\left[ {\because \cos \left( { - \theta } \right) = \cos \theta } \right]$
Now take common the common terms we have,
$ \Rightarrow \sin 2x\left( {2\cos x + 3} \right) = \cos 2x\left( {2\cos x + 3} \right)$
$ \Rightarrow \left( {2\cos x + 3} \right)\left( {\sin 2x - \cos 2x} \right) = 0$………………. (1)
$ \Rightarrow 2\cos x + 3 = 0$
$ \Rightarrow \cos x = \dfrac{{ - 3}}{2}$ (Which is not possible as the value of cosine is always lie between -1 to 1)
Again from equation (1) we have,
$ \Rightarrow \left( {\sin 2x - \cos 2x} \right) = 0$
$ \Rightarrow \sin 2x = \cos 2x$
$ \Rightarrow \dfrac{{\sin 2x}}{{\cos 2x}} = 1$
$ \Rightarrow \tan 2x = 1 = \tan \left( {n\pi + \dfrac{\pi }{4}} \right)$ (General solution where n = 0, 1, 2, 3, …………………. )
So on comparing
$ \Rightarrow 2x = n\pi + \dfrac{\pi }{4}$
$ \Rightarrow x = n\dfrac{\pi }{2} + \dfrac{\pi }{8}$
Now, we have to find out the value of x in the interval$0 \leqslant x \leqslant 2\pi $.
For n = 0
$ \Rightarrow x = \dfrac{\pi }{8}$.
For n = 1
$ \Rightarrow x = \dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{5\pi }}{8}$
For n = 2
$ \Rightarrow x = 2\dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{9\pi }}{8}$
For n = 3
$ \Rightarrow x = 3\dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{13\pi }}{8}$
For n = 4
$ \Rightarrow x = 4\dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{17\pi }}{8} > 2\pi $ (So this solution is not considered)
Hence the required solution is $x = \dfrac{\pi }{8},\dfrac{{5\pi }}{8},\dfrac{{9\pi }}{8},\dfrac{{13\pi }}{8}$
Hence option (b) is correct.
Note: Whenever we face such types of problems the key concept is firstly to reach to the value of x but application of various trigonometric and algebraic identities. As the trigonometric functions involved are mostly periodic in nature therefore the solution also varies in different intervals so we need to keep in mind about the interval in which the solution is being asked.
Complete step-by-step answer:
Given equation is
$\sin x + 3\sin 2x + \sin 3x = \cos x + 3\cos 2x + \cos 3x$
$ \Rightarrow \sin x + \sin 3x + 3\sin 2x = \cos x + \cos 3x + 3\cos 2x$
We have to find out the solution of this equation in the interval\[0 \leqslant x \leqslant 2\pi \].
Now as we know
$
\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) \\
\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) \\
$
So, use this property in the above equation we have,
$ \Rightarrow \sin x + \sin 3x + 3\sin 2x = \cos x + \cos 3x + 3\cos 2x$
$ \Rightarrow 2\sin \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right) + 3\sin 2x = 2\cos \left( {\dfrac{{x + 3x}}{2}} \right)\cos \left( {\dfrac{{x - 3x}}{2}} \right) + 3\cos 2x$
Now simplify the above equation we have,
$ \Rightarrow 2\sin 2x\cos x + 3\sin 2x = 2\cos 2x\cos x + 3\cos 2x$ $\left[ {\because \cos \left( { - \theta } \right) = \cos \theta } \right]$
Now take common the common terms we have,
$ \Rightarrow \sin 2x\left( {2\cos x + 3} \right) = \cos 2x\left( {2\cos x + 3} \right)$
$ \Rightarrow \left( {2\cos x + 3} \right)\left( {\sin 2x - \cos 2x} \right) = 0$………………. (1)
$ \Rightarrow 2\cos x + 3 = 0$
$ \Rightarrow \cos x = \dfrac{{ - 3}}{2}$ (Which is not possible as the value of cosine is always lie between -1 to 1)
Again from equation (1) we have,
$ \Rightarrow \left( {\sin 2x - \cos 2x} \right) = 0$
$ \Rightarrow \sin 2x = \cos 2x$
$ \Rightarrow \dfrac{{\sin 2x}}{{\cos 2x}} = 1$
$ \Rightarrow \tan 2x = 1 = \tan \left( {n\pi + \dfrac{\pi }{4}} \right)$ (General solution where n = 0, 1, 2, 3, …………………. )
So on comparing
$ \Rightarrow 2x = n\pi + \dfrac{\pi }{4}$
$ \Rightarrow x = n\dfrac{\pi }{2} + \dfrac{\pi }{8}$
Now, we have to find out the value of x in the interval$0 \leqslant x \leqslant 2\pi $.
For n = 0
$ \Rightarrow x = \dfrac{\pi }{8}$.
For n = 1
$ \Rightarrow x = \dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{5\pi }}{8}$
For n = 2
$ \Rightarrow x = 2\dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{9\pi }}{8}$
For n = 3
$ \Rightarrow x = 3\dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{13\pi }}{8}$
For n = 4
$ \Rightarrow x = 4\dfrac{\pi }{2} + \dfrac{\pi }{8} = \dfrac{{17\pi }}{8} > 2\pi $ (So this solution is not considered)
Hence the required solution is $x = \dfrac{\pi }{8},\dfrac{{5\pi }}{8},\dfrac{{9\pi }}{8},\dfrac{{13\pi }}{8}$
Hence option (b) is correct.
Note: Whenever we face such types of problems the key concept is firstly to reach to the value of x but application of various trigonometric and algebraic identities. As the trigonometric functions involved are mostly periodic in nature therefore the solution also varies in different intervals so we need to keep in mind about the interval in which the solution is being asked.
Last updated date: 25th Sep 2023
•
Total views: 360.9k
•
Views today: 7.60k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Summary of the poem Where the Mind is Without Fear class 8 english CBSE

The poet says Beauty is heard in Can you hear beauty class 6 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

What is the past tense of read class 10 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
