This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Step 1: Rewrite the trigonometric functions in terms of (x) and (x). We know that (2x) = 2(x)(x) and (x) = ((x))/((x)). Substitute these into the given equation: 2(x)(x) = ((x))/((x)) Note that for (x) to be defined, (x) ≠ 0. This means x ≠ n for any integer n. Step 2: Rearrange the equation to solve for x. Multiply both sides by (x) (which is non-zero based on the restriction): 2^2(x)(x) = (x) Move all terms to one side: 2^2(x)(x) - (x) = 0 Step 3: Factor out the common term (x). (x)(2^2(x) - 1) = 0 Step 4: Set each factor equal to zero and solve for x. This gives two separate cases: Case 1: (x) = 0 Case 2: 2^2(x) - 1 = 0 Step 5: Solve Case 1. If (x) = 0, then x is an odd multiple of ()/(2). x = ()/(2) + n where n is an integer. Step 6: Solve Case 2. If 2^2(x) - 1 = 0, then: 2^2(x) = 1 ^2(x) = (1)/(2) We can use the identity (2x) = 1 - 2^2(x). So, 2^2(x) - 1 = -(1 - 2^2(x)) = -(2x). Therefore, the equation becomes -(2x) = 0, or (2x) = 0. If (2x) = 0, then 2x is an odd multiple of ()/(2). 2x = ()/(2) + n Divide by 2: x = ()/(4) + (n)/(2) where n is an integer. Step 7: Verify solutions against the restriction (x) ≠ 0. For x = ()/(2) + n, (x) = ± 1, which is not zero. These solutions are valid. For x = ()/(4) + (n)/(2), (x) = ± sqrt(2)2, which is not zero. These solutions are valid. The general solutions are the union of the solutions from Case 1 and Case 2. The solutions are: x = ()/(2) + n or x = ()/(4) + (n)/(2) where n Z. The final answer is x = ()/(2) + n or x = ()/(4) + (n)/(2), where n Z. Send me the next one 📸