2sin(2x+π/3) - 3cos x = sin 2x - √3
2(sin 2x·1/2 + cos 2x·√3/2) - 3cos x = sin 2x - √3
√3cos 2x - 3cos x = - √3
2√3cos²x - √3 - 3cos x = - √3
2√3cos x·(cos x - √3/2) = 0
cos x = 0
x = π/2 + πn, n ∈ Z
cos x - √3/2 = 0
cos x = √3/2
x = (+/-) π/6 + 2πk, k ∈ Z
Ответ: x = π/2 + πn, x = (+/-) π/6 + 2πk; n, k ∈ Z.
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Answers & Comments
2sin(2x+π/3) - 3cos x = sin 2x - √3
2(sin 2x·cos π/3 + cos 2x·sin π/3) - 3cos x = sin 2x - √32(sin 2x·1/2 + cos 2x·√3/2) - 3cos x = sin 2x - √3
sin 2x + √3cos 2x - 3cos x = sin 2x - √3√3cos 2x - 3cos x = - √3
√3(2cos²x - 1) - 3cos x = - √32√3cos²x - √3 - 3cos x = - √3
2√3cos²x - 3cos x = 02√3cos x·(cos x - √3/2) = 0
cos x = 0
x = π/2 + πn, n ∈ Z
cos x - √3/2 = 0
cos x = √3/2
x = (+/-) π/6 + 2πk, k ∈ Z
Ответ: x = π/2 + πn, x = (+/-) π/6 + 2πk; n, k ∈ Z.