搜搜做题作业网证明以下三角函数等式state,with reasons,which of the followings equatio
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证明以下三角函数等式
state,with reasons,which of the followings equations have no roots.
用理由说明（证明） 以下的等式 不成立（无根,无值）
1）2sinθ=3 2)sinθ+cosθ=0 3)sinθ+cosθ=2
4)4cosec^2θ-1=0 5)cosecθ=sinθ 6)secθ=sinθ
请详细解释下 谢谢
state,with reasons,which of the followings equations have no roots.
用理由说明（证明） 以下的等式 不成立（无根,无值）
1）2sinθ=3 2)sinθ+cosθ=0 3)sinθ+cosθ=2
4)4cosec^2θ-1=0 5)cosecθ=sinθ 6)secθ=sinθ
请详细解释下 谢谢
1,
sin(x) = 3/2 > 1.has no root.
2,
tan(x) = -1,x = kPI - PI/4,has roots.
3,
2 = 2^(1/2)sin[x + PI/4],
sin[x + PI/4] = 2^(1/2) > 1.has no root.
4,
cosec(x) = 1/sin(x).
1 = 4[1/sin(x)]^2,[sin(x)]^2 = 4 > 1,has no root.
5,
[sin(x)]^2 = 1,x = kPI + PI/2,has roots.
6,
1 = sin(x)cos(x) = [sin(2x)]/2,
sin(2x) = 2 > 1,has no root.
sin(x) = 3/2 > 1.has no root.
2,
tan(x) = -1,x = kPI - PI/4,has roots.
3,
2 = 2^(1/2)sin[x + PI/4],
sin[x + PI/4] = 2^(1/2) > 1.has no root.
4,
cosec(x) = 1/sin(x).
1 = 4[1/sin(x)]^2,[sin(x)]^2 = 4 > 1,has no root.
5,
[sin(x)]^2 = 1,x = kPI + PI/2,has roots.
6,
1 = sin(x)cos(x) = [sin(2x)]/2,
sin(2x) = 2 > 1,has no root.
证明以下三角函数等式state,with reasons,which of the followings equatio
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