9514 1404 393
Answer:
x ∈ {π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6}
Explanation:
We can divide by 6 and use some identities to rewrite this as ...
2tan(x)/(1-tan(x)²) -1/tan(x) = 0
2tan(x)² -(1 -tan(x)²) = 0 . . . . . . . . . multiply* by tan(x)(1 -tan(x)²)
3tan(x)² = 1
tan(x) = ±1/√3 ⇒ x = {π/6, 5π/6, 7π/6, 11π/6}
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Interestingly, if we use different identities, we can rewrite it as ...
2cot(x)/(cot(x)² -1) -cot(x) = 0
cot(x)(3 -cot(x)²) = 0 . . . . . . . multiply by (cot(x)² -1)
In addition to the zeros found above, this equation also has solutions where cot(x) = 0. That is ...
x = {π/2, 3π/2}
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Usually, the first method shown would give all the solutions. Fortunately, we had a graph of the function, so we could see that the first set of solutions found was not complete. (This is one reason I like to use a graphing calculator for guidance on problems like this.)
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* In the first solution, we multiplied by the denominator tan(x)(1 -tan(x)²). It was multiplying by tan(x) that made the π/2, 3π/2 solutions disappear. Had we kept 1/tan(x) as a numerator factor of cot(x), we would have had the same result as seen in the second approach.