Đề bài
Giải phương trình:
a) (sin left( {2x – frac{pi }{6}} right) = – frac{1}{2})
b) (sin left( {frac{x}{3} + frac{pi }{2}} right) = frac{{sqrt 3 }}{2})
c) (cos left( {2x + frac{pi }{5}} right) = frac{{sqrt 2 }}{2})
d) (2cos frac{x}{2} + sqrt 3 = 0)
e) (sqrt 3 tan left( {2x + frac{pi }{3}} right) – 1 = 0)
g) (cot left( {3x + pi } right) = – 1)
Lời giải chi tiết
a) Ta có (sin left( { – frac{pi }{6}} right) = – frac{1}{2}), phương trình trở thành:
(sin left( {2x – frac{pi }{6}} right) = sin left( { – frac{pi }{6}} right) Leftrightarrow left[ begin{array}{l}2x – frac{pi }{6} = – frac{pi }{6} + k2pi 2x – frac{pi }{6} = pi + frac{pi }{6} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}2x = k2pi 2x = frac{{4pi }}{3} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}x = kpi x = frac{{2pi }}{3} + kpi end{array} right.left( {k in mathbb{Z}} right))
b) Ta có (sin frac{pi }{3} = frac{{sqrt 3 }}{2}), phương trình trở thành:
(sin left( {frac{x}{3} + frac{pi }{2}} right) = sin frac{pi }{3} Leftrightarrow left[ begin{array}{l} + frac{pi }{2} = frac{pi }{3} + k2pi frac{x}{3} + frac{pi }{2} = pi – frac{pi }{3} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}frac{x}{3} = – frac{pi }{6} + k2pi frac{x}{3} = frac{pi }{6} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}x = – frac{pi }{2} + k6pi x = frac{pi }{2} + k6pi end{array} right.left( {k in mathbb{Z}} right))
c) Ta có (cos frac{pi }{4} = frac{{sqrt 2 }}{2}), phương trình trở thành:
(cos left( {2x + frac{pi }{5}} right) = cos frac{pi }{4} Leftrightarrow left[ begin{array}{l}2x + frac{pi }{5} = frac{pi }{4} + k2pi 2x + frac{pi }{5} = – frac{pi }{4} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}2x = frac{pi }{{20}} + k2pi 2x = – frac{{9pi }}{{20}} + k2pi end{array} right.left[ begin{array}{l}x = frac{pi }{{40}} + kpi x = – frac{{9pi }}{{40}} + kpi end{array} right.left( {k in mathbb{Z}} right))
d) (2cos frac{x}{2} + sqrt 3 = 0 Leftrightarrow cos frac{x}{2} = – frac{{sqrt 3 }}{2})
Ta có (cos frac{{5pi }}{6} = – frac{{sqrt 3 }}{2}), phương trình trở thành:
(cos frac{x}{2} = cos frac{{5pi }}{6} Leftrightarrow left[ begin{array}{l}frac{x}{2} = frac{{5pi }}{6} + k2pi frac{x}{2} = – frac{{5pi }}{6} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}x = frac{{5pi }}{3} + k4pi x = – frac{{5pi }}{3} + k4pi end{array} right.left( {k in mathbb{Z}} right))
e) (sqrt 3 tan left( {2x + frac{pi }{3}} right) – 1 = 0 Leftrightarrow tan left( {2x + frac{pi }{3}} right) = frac{1}{{sqrt 3 }})
Ta có (tan frac{pi }{6} = frac{1}{{sqrt 3 }}), phương trình trở thành:
(tan left( {2x + frac{pi }{3}} right) = tan frac{pi }{6} Leftrightarrow 2x + frac{pi }{3} = frac{pi }{6} + kpi Leftrightarrow 2x = – frac{pi }{6} + kpi Leftrightarrow x = – frac{pi }{12} + kfrac{pi }{2}left( {k in mathbb{Z}} right))
f) Ta có (cot left( { – frac{pi }{4}} right) = – 1), phương trình trở thành:
(cot left( {3x + pi } right) = cot frac{{ – pi }}{4} Leftrightarrow 3x + pi = frac{{ – pi }}{4} + kpi Leftrightarrow 3x = frac{{ – 5pi }}{4} + kpi Leftrightarrow x = frac{{ – 5pi }}{{12}} + kfrac{pi }{3}left( {k in mathbb{Z}} right))
