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Answer 1

$a, 2 \sin 3 x - \sin 2 x + \sqrt{3} \cos 2 x = 0 < = > \sin 2 x - \sqrt{3} \cos 2 x = 2 \sin 3 x < = > \frac{1}{2} \sin 2 x - \frac{\sqrt{3}}{2} \cos 2 x = \sin 3 x < = > \sin 2 x . \cos \frac{π}{3} - \cos 2 x . \sin \frac{π}{3} = \sin 3 x < = > \sin (2 x - \frac{π}{3}) = \sin 3 x < = > 2 x - \frac{π}{3} = 3 x + k 2 π hoặc 2 x - \frac{π}{3} = π - 3 x + k 2 π, k \in Z < = > - x = \frac{π}{3} + k 2 π hoặc 5 x = \frac{4 π}{3} + k 2 π, k \in Z < = > x = - \frac{π}{3} + k 2 π hoặc x = \frac{4 π}{15} + \frac{k 2 π}{3}, k \in Z Vậy S = \{- \frac{π}{3} + k 2 π; \frac{4 π}{15} + \frac{k 2 π}{3}\}, k \in Z$

$b, 3 \sin 2 x + 4 \cos 2 x + 5 \cos 2003 x = 0 < = > 4 \cos 2 x + 3 \sin 2 x = - 5 \cos 2003 x < = > \frac{4}{5} . \cos 2 x + \frac{3}{5} \sin 2 x = \cos (2003 x + π) < = > \cos 2 x . \cos a + \sin 2 x . \sin a = \cos (2003 + π) Với a là góc có cosa = \frac{4}{5}; sina = \frac{3}{5} < = > \cos (2 x - a) = \cos (2003 x + π) < = > 2 x - a = 2003 x + π + k 2 π hoặc 2 x - a = - 2003 x - π + k 2 π, k \in Z < = > - 2001 x = a + π + k 2 π hoặc 2005 x = a - π + k 2 π, k \in Z < = > x = - \frac{1}{2001} - \frac{π}{2001} + \frac{k 2 π}{2001} hoặc x = \frac{1}{2005} - \frac{π}{2005} + \frac{k 2 π}{2005}, k \in Z vậy S = \{- \frac{1}{2001} - \frac{π}{2001} + \frac{k 2 π}{2001}; \frac{1}{2005} - \frac{π}{2005} + \frac{k 2 π}{2005}\}, k \in Z$

$c, \sqrt{3} \sin 4 x - \cos 4 x = \sin x - \sqrt{3} \cos x < = > \frac{\sqrt{3}}{2} \sin 4 x - \frac{1}{2} \cos 4 x = \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x < = > \sin 4 x . \cos \frac{π}{6} - \cos 4 x . \sin \frac{π}{6} = \sin x . \cos \frac{π}{3} - \cos x . \sin \frac{π}{3} < = > \sin (4 x - \frac{π}{6}) = \sin (x - \frac{π}{3}) < = > 4 x - \frac{π}{6} = x - \frac{π}{3} + k 2 π hoặc 4 x - \frac{π}{6} = π - x + \frac{π}{3} + k 2 π, k \in Z < = > 3 x = \frac{- π}{6} + k 2 π hoặc 5 x = \frac{3 π}{2} + k 2 π, k \in Z < = > x = - \frac{π}{18} + \frac{k 2 π}{3} hoặc x = \frac{3 π}{10} + \frac{k 2 π}{5}, k \in Z Vậy S = \{- \frac{π}{18} + \frac{k 2 π}{3}; \frac{3 π}{10} + \frac{k 2 π}{5}\}, k \in Z$ .

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Answer 2

$a, 2 \sin 3 x - \sin 2 x + \sqrt{3} \cos 2 x = 0 < = > \sin 2 x - \sqrt{3} \cos 2 x = 2 \sin 3 x < = > \frac{1}{2} \sin 2 x - \frac{\sqrt{3}}{2} \cos 2 x = \sin 3 x < = > \sin 2 x . \cos \frac{π}{3} - \cos 2 x . \sin \frac{π}{3} = \sin 3 x < = > \sin (2 x - \frac{π}{3}) = \sin 3 x < = > 2 x - \frac{π}{3} = 3 x + k 2 π hoặc 2 x - \frac{π}{3} = π - 3 x + k 2 π, k \in Z < = > - x = \frac{π}{3} + k 2 π hoặc 5 x = \frac{4 π}{3} + k 2 π, k \in Z < = > x = - \frac{π}{3} + k 2 π hoặc x = \frac{4 π}{15} + \frac{k 2 π}{3}, k \in Z Vậy S = \{- \frac{π}{3} + k 2 π; \frac{4 π}{15} + \frac{k 2 π}{3}\}, k \in Z$

$b, 3 \sin 2 x + 4 \cos 2 x + 5 \cos 2003 x = 0 < = > 4 \cos 2 x + 3 \sin 2 x = - 5 \cos 2003 x < = > \frac{4}{5} . \cos 2 x + \frac{3}{5} \sin 2 x = \cos (2003 x + π) < = > \cos 2 x . \cos a + \sin 2 x . \sin a = \cos (2003 + π) Với a là góc có cosa = \frac{4}{5}; sina = \frac{3}{5} < = > \cos (2 x - a) = \cos (2003 x + π) < = > 2 x - a = 2003 x + π + k 2 π hoặc 2 x - a = - 2003 x - π + k 2 π, k \in Z < = > - 2001 x = a + π + k 2 π hoặc 2005 x = a - π + k 2 π, k \in Z < = > x = - \frac{1}{2001} - \frac{π}{2001} + \frac{k 2 π}{2001} hoặc x = \frac{1}{2005} - \frac{π}{2005} + \frac{k 2 π}{2005}, k \in Z vậy S = \{- \frac{1}{2001} - \frac{π}{2001} + \frac{k 2 π}{2001}; \frac{1}{2005} - \frac{π}{2005} + \frac{k 2 π}{2005}\}, k \in Z$

$c, \sqrt{3} \sin 4 x - \cos 4 x = \sin x - \sqrt{3} \cos x < = > \frac{\sqrt{3}}{2} \sin 4 x - \frac{1}{2} \cos 4 x = \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x < = > \sin 4 x . \cos \frac{π}{6} - \cos 4 x . \sin \frac{π}{6} = \sin x . \cos \frac{π}{3} - \cos x . \sin \frac{π}{3} < = > \sin (4 x - \frac{π}{6}) = \sin (x - \frac{π}{3}) < = > 4 x - \frac{π}{6} = x - \frac{π}{3} + k 2 π hoặc 4 x - \frac{π}{6} = π - x + \frac{π}{3} + k 2 π, k \in Z < = > 3 x = \frac{- π}{6} + k 2 π hoặc 5 x = \frac{3 π}{2} + k 2 π, k \in Z < = > x = - \frac{π}{18} + \frac{k 2 π}{3} hoặc x = \frac{3 π}{10} + \frac{k 2 π}{5}, k \in Z Vậy S = \{- \frac{π}{18} + \frac{k 2 π}{3}; \frac{3 π}{10} + \frac{k 2 π}{5}\}, k \in Z$ .

Answer 3

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