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JEE Main Maths Trigonometry Previous Year Questions With Solutions
**Question 1:**
The general solution of sin x - 3 sin2x + sin3x = cos x - 3 cos2x + cos3x is
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Welcome to this JEE Main trigonometry problem. We need to solve the equation sin x minus 3 sin 2x plus sin 3x equals cos x minus 3 cos 2x plus cos 3x. This is a complex trigonometric equation that requires careful algebraic manipulation and the use of trigonometric identities. Let's visualize the basic sine and cosine functions first, as they form the foundation of our solution.
The first step in solving this trigonometric equation is to rearrange the terms strategically. We group the sine terms and cosine terms that can be combined using sum-to-product formulas. On the left side, we group sin x plus sin 3x, and on the right side, we group cos x plus cos 3x. This rearrangement sets us up to apply trigonometric identities in the next step.
Now we apply the sum-to-product formulas. For sine A plus sine B, we get 2 sine of A plus B over 2 times cosine of A minus B over 2. Similarly for cosine terms. Applying this to sin x plus sin 3x gives us 2 sine 2x cosine x. For cos x plus cos 3x, we get 2 cosine 2x cosine x. Our equation becomes 2 sine 2x cosine x minus 3 sine 2x equals 2 cosine 2x cosine x minus 3 cosine 2x.
Now we factor and solve the equation. We can factor out the common term sin 2x minus cos 2x, giving us the product of two factors equal to zero. This gives us two cases to consider. Case 1: sin 2x equals cos 2x, which means tan 2x equals 1. This gives us 2x equals n pi plus pi over 4, so x equals n pi over 2 plus pi over 8. Case 2: cos x equals 3 over 2, which is impossible since cosine values must be between negative 1 and 1. Therefore, our general solution is x equals n pi over 2 plus pi over 8, where n is any integer.
Here is our final answer. The general solution is x equals n pi over 2 plus pi over 8, where n is any integer. This gives us an infinite set of solutions that are evenly spaced. The first few solutions are pi over 8, 5 pi over 8, 9 pi over 8, and so on. The solutions repeat every pi over 2 units, creating a regular pattern. This completes our solution to the JEE Main trigonometry problem.