Đề bài
Giải phương trình
a) (sin left( {2x + frac{pi }{4}} right) = sin x)
b) (sin 2x = cos 3x)
c) ({cos ^2}2x = {cos ^2}left( {x + frac{pi }{6}} right))
Lời giải chi tiết
a)
(sin left( {2x + frac{pi }{4}} right) = sin x Leftrightarrow left[ begin{array}{l}2x + frac{pi }{4} = x + k2pi 2x + frac{pi }{4} = pi – x + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}x = – frac{pi }{4} + k2pi 3x = pi – frac{pi }{4} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}x = – frac{pi }{4} + k2pi x = frac{pi }{4} + frac{{k2pi }}{3}end{array} right.;k in Z)
b) Ta có (sin 2x = cos left( {frac{pi }{2} – 2x} right)) (quan hệ giữa giá trị lượng giác của hai góc phụ nhau).
(begin{array}{l}sin 2x = cos 3x Leftrightarrow cos 3x = cos left( {frac{pi }{2} – 2x} right) Leftrightarrow left[ begin{array}{l}3x = frac{pi }{2} – 2x + k2pi 3x = – left( {frac{pi }{2} – 2x} right) + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}5x = frac{pi }{2} + k2pi x = – frac{pi }{2} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}x = frac{pi }{{10}} + frac{{k2pi }}{5}x = – frac{pi }{2} + k2pi end{array} right.end{array})
c)
(begin{array}{l}{cos ^2}2x = {cos ^2}left( {x + frac{pi }{6}} right) Leftrightarrow left[ begin{array}{l}cos 2x = cos left( {x + frac{pi }{6}} right)cos 2x = – cos left( {x + frac{pi }{6}} right)end{array} right. Leftrightarrow left[ begin{array}{l}cos 2x = cos left( {x + frac{pi }{6}} right)cos 2x = cos left( {pi – left( {x + frac{pi }{6}} right)} right)end{array} right. Leftrightarrow left[ begin{array}{l}cos 2x = cos left( {x + frac{pi }{6}} right)cos 2x = cos left( {frac{{5pi }}{6} – x} right)end{array} right.end{array})
Với (cos 2x = cos left( {x + frac{pi }{6}} right) Leftrightarrow left[ begin{array}{l}2x = – left( {x + frac{pi }{6}} right) + k2pi 2x = x + frac{pi }{6} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}3x = – frac{pi }{6} + k2pi x = frac{pi }{6} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}x = – frac{pi }{{18}} + frac{{k2pi }}{3}x = frac{pi }{6} + k2pi end{array} right.)
Với (cos 2x = cos left( {frac{{5pi }}{6} – x} right) Leftrightarrow left[ begin{array}{l}2x = frac{{5pi }}{6} – x + k2pi 2x = – left( {frac{{5pi }}{6} – x} right) + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}3x = frac{{5pi }}{6} + k2pi x = – frac{{5pi }}{6} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}x = frac{{5pi }}{{18}} + frac{{k2pi }}{3}x = – frac{{5pi }}{6} + k2pi end{array} right.)
