求直线y= 2sin(2x+π/6)+2sin(2x-π/3)与直线y=根号6的图像在(0,π)上所在交点的坐标
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求直线y= 2sin(2x+π/6)+2sin(2x-π/3)与直线y=根号6的图像在(0,π)上所在交点的坐标
![求直线y= 2sin(2x+π/6)+2sin(2x-π/3)与直线y=根号6的图像在(0,π)上所在交点的坐标](/uploads/image/z/19241336-56-6.jpg?t=%E6%B1%82%E7%9B%B4%E7%BA%BFy%3D+2sin%282x%2B%CF%80%2F6%29%2B2sin%282x-%CF%80%2F3%29%E4%B8%8E%E7%9B%B4%E7%BA%BFy%3D%E6%A0%B9%E5%8F%B76%E7%9A%84%E5%9B%BE%E5%83%8F%E5%9C%A8%EF%BC%880%2C%CF%80%EF%BC%89%E4%B8%8A%E6%89%80%E5%9C%A8%E4%BA%A4%E7%82%B9%E7%9A%84%E5%9D%90%E6%A0%87)
y=2sin(2x+π/6)+2sin(2x-π/3)
=2sin(2x-π/3+π/2)+2sin(2x-π/3)
=2cos(2x-π/3)+2sin(2x-π/3)
=2√2[sin(2x-π/3)cos(π/4)+cos(2x-π/3)sin(π/4)]
=2√2sin(2x-π/3+π/4)=√6
所以sin(2x-π/12)=√3/2=sin(π/3)
∵x∈(0,π)
∴2x-π/12∈(-π/12,2π-π/12)
∴2x1-π/12=π/3,2x2-π/12=2π/3,
∴x1=5π/24,x2=9π/24
所以交点有2个,分别是(5π/24,√6),(9π/24,√6)
再问: 谢谢喽
=2sin(2x-π/3+π/2)+2sin(2x-π/3)
=2cos(2x-π/3)+2sin(2x-π/3)
=2√2[sin(2x-π/3)cos(π/4)+cos(2x-π/3)sin(π/4)]
=2√2sin(2x-π/3+π/4)=√6
所以sin(2x-π/12)=√3/2=sin(π/3)
∵x∈(0,π)
∴2x-π/12∈(-π/12,2π-π/12)
∴2x1-π/12=π/3,2x2-π/12=2π/3,
∴x1=5π/24,x2=9π/24
所以交点有2个,分别是(5π/24,√6),(9π/24,√6)
再问: 谢谢喽
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