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For x ∈( 0,p ) , the equation sinx + 2sin 2x – sin 3x = 3 has (JEE Adv. 2014)
- a)infinitely many solutions
- b)three solutions
- c)one solution
- d)no solution
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
For x ∈( 0,p ) , the equation sinx + 2sin 2x – sin 3x = 3 h...
sin x + 2sin 2x – sin 3x = 3
⇒ sin x + 4sin x cos x – 3sin x + 4sin3 x = 3
⇒ sin x (–2 + 2cos x + 4sin2 x) = 3
⇒ sin x (– 2 + 2cos x + 4 – 4cos2 x) = 3
⇒ sin x + 4sin x cos x – 3sin x + 4sin3 x = 3
⇒ sin x (–2 + 2cos x + 4sin2 x) = 3
⇒ sin x (– 2 + 2cos x + 4 – 4cos2 x) = 3
⇒ 2 + 2 cos x – 
⇒
⇒ 
LHS 
and RHS ≥ 3
and RHS ≥ 3∴ Equation has no solution.
Most Upvoted Answer
For x ∈( 0,p ) , the equation sinx + 2sin 2x – sin 3x = 3 h...
Understanding the Equation
The equation given is:
\[ \sin x + 2\sin 2x - \sin 3x = 3 \]
We need to assess the possibility of solutions for this equation in the interval \( x \in (0, \pi) \).
Range of Sinusoidal Functions
- The sine function, \( \sin x \), oscillates between -1 and 1.
- Therefore, \( \sin 3x \), \( \sin 2x \), and \( \sin x \) also have the same range of [-1, 1].
Maximum Value of the Left-Hand Side
Now, consider the left-hand side of the equation:
\[ \sin x + 2\sin 2x - \sin 3x \]
- The maximum possible value of \( \sin x \) is 1.
- The maximum possible value of \( \sin 2x \) is also 1, thus \( 2\sin 2x \) can reach 2.
- The minimum possible value of \( -\sin 3x \) is -1.
Combining these maximums:
\[ 1 + 2 + 1 = 4 \]
However, since we only consider the maximum contributions of each term:
\[ \text{Maximum value} = 1 + 2 - (-1) = 4 - 1 = 3 \]
Conclusion
Despite the maximum value being 4, we need to analyze if the expression can actually equal 3:
- The equation \( \sin x + 2\sin 2x - \sin 3x = 3 \) implies that the left-hand side must equal exactly 3.
- This is only achievable if each component of the left-hand side reaches its maximum possible value, which is not feasible simultaneously due to the periodic and oscillatory nature of sine functions.
Thus, there are no values of \( x \) in the interval \( (0, \pi) \) for which this equation holds true.
Final Answer
Hence, the equation has no solution in the given interval, confirming option 'D'.
The equation given is:
\[ \sin x + 2\sin 2x - \sin 3x = 3 \]
We need to assess the possibility of solutions for this equation in the interval \( x \in (0, \pi) \).
Range of Sinusoidal Functions
- The sine function, \( \sin x \), oscillates between -1 and 1.
- Therefore, \( \sin 3x \), \( \sin 2x \), and \( \sin x \) also have the same range of [-1, 1].
Maximum Value of the Left-Hand Side
Now, consider the left-hand side of the equation:
\[ \sin x + 2\sin 2x - \sin 3x \]
- The maximum possible value of \( \sin x \) is 1.
- The maximum possible value of \( \sin 2x \) is also 1, thus \( 2\sin 2x \) can reach 2.
- The minimum possible value of \( -\sin 3x \) is -1.
Combining these maximums:
\[ 1 + 2 + 1 = 4 \]
However, since we only consider the maximum contributions of each term:
\[ \text{Maximum value} = 1 + 2 - (-1) = 4 - 1 = 3 \]
Conclusion
Despite the maximum value being 4, we need to analyze if the expression can actually equal 3:
- The equation \( \sin x + 2\sin 2x - \sin 3x = 3 \) implies that the left-hand side must equal exactly 3.
- This is only achievable if each component of the left-hand side reaches its maximum possible value, which is not feasible simultaneously due to the periodic and oscillatory nature of sine functions.
Thus, there are no values of \( x \) in the interval \( (0, \pi) \) for which this equation holds true.
Final Answer
Hence, the equation has no solution in the given interval, confirming option 'D'.
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For x ∈( 0,p ) , the equation sinx + 2sin 2x – sin 3x = 3 has (JEE Adv. 2014)a)infinitely many solutionsb)three solutionsc)one solutiond)no solutionCorrect answer is option 'D'. Can you explain this answer?
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