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[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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Chào các bạn,
Mình có 1 số bài lượng giác làm không ra, nay mình post lên đây để mọi người cùng tham khảo, làm va học tập, nhờ các bạn giúp đỡ:
[TEX]1) 3\cos x + 2|\sin x| = k;k = 2;k = 3\\2)\frac{{1 - \cos 4x}}{{2\sin 2x}} = \frac{{\sin 4x}}{{1 + \cos 4x}}[/TEX]
[TEX]\\ 3){\rm{ }}6tgx{\rm{ }} + {\rm{ }}a\cot g3x{\rm{ }} = {\rm{ }}tg2x.{\rm{ }}Giai{\rm{ }}khi{\rm{ }}a{\rm{ }} = {\rm{ }}0{\rm{ }}va{\rm{ }}a{\rm{ }} = {\rm{ }}5 \\4){\rm{ }}|\sin x{\rm{ }} - {\rm{ }}\cos x|{\rm{ }} + {\rm{ }}4\sin 2x{\rm{ }} = {\rm{ }}1 [/TEX]
[TEX]\\5){\rm{ }}t{g^2}x = \frac{{1 - {{\cos }^3}x}}{{1 - {{\sin }^3}x}} \\6){\rm{ }}t{g^2}x = \frac{{1 - \cos |x|}}{{1 - \sin |x|}} \\7){\rm{ }}\cos x{\rm{ }} + {\rm{ }}\frac{1}{{\cos x}} + \sin x + \frac{1}{{\sin x}} = \frac{{10}}{3} \\8){\rm{ }}3\cos x{\rm{ }} + {\rm{ }}4\sin x{\rm{ }} + {\rm{ }}\frac{6}{{3\cos x{\rm{ }} + {\rm{ }}4\sin x + 1}} = 6 [/TEX]
[TEX]\\9){\rm{ }}4{\cos ^2}x + 3t{g^2}x - 4\sqrt 3 \cos x + 2\sqrt 3 tgx + 4 = 0 \\10)3tg3x + \cot g2x = 2tgx + \frac{2}{{\sin 4x}} \\11)tg({120^0} + 3x) - tg({140^0} - x) = 2\sin ({80^0} + 2x) \\12)2(tgx - \sin x) + 3(\cot gx - \cos x) + 5 = 0 \\13){\sin ^3}x(1 + \cot gx) + {\cos ^3}x(1 + tgx) = 2\sqrt {\sin x\cos x} \\14)\frac{{{{\sin }^3}x + {{\cos }^3}x}}{{2\cos x - \sin x}} = \cos 2x \\15)\cos 2x - t{g^2}x = \frac{{{{\cos }^2}x - {{\cos }^3}x - 1}}{{{{\cos }^2}x}},x \in [1;70]rad \\16){\sin ^3}x + {\cos ^3}x = 2 - {\sin ^4}x \\17){\cos ^4}x - {\sin ^4}x = |\cos x| + |\sin x| \\18){\sin ^2}x + {\sin ^2}y + {\sin ^2}(x + y) = \frac{9}{4} \\19)2\sin (3x + \frac{\pi }{4}) = \sqrt {1 + 8\sin 2x{{\cos }^2}2x} \\20)tgx + t{g^2}x + t{g^3}x + \cot gx + \cot {g^2}x + \cot {g^3}x = 6 \\21)\cot g2x + \cot g3x + \frac{1}{{\sin x\sin 2x\sin 3x}} = 0 \\22){(tgx + \frac{1}{4}\cot gx)^n} = {\cos ^n}x + {\sin ^n}x,(n = 2,3,4...) \\23)\frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{{{\sin }^6}x + {{\cos }^6}x}}{{4{{\cos }^2}2x + {{\sin }^2}2x}} \\24)\cos 2x - \cos 6x + 4(3\sin x - 4{\sin ^3}x + 1) = 0 [/TEX]
[TEX]\\25){\left( {{{\sin }^3}\frac{x}{2} + \frac{1}{{{{\sin }^3}\frac{x}{2}}}} \right)^2} + {\left( {{{\cos }^3}\frac{x}{2} + \frac{1}{{{{\cos }^3}\frac{x}{2}}}} \right)^2} = \frac{{81}}{4}{\cos ^2}4x \\26)\frac{{{{\sin }^4}\frac{x}{2} + {{\cos }^4}\frac{x}{2}}}{{1 - \sin x}} - t{g^2}x\sin x = \frac{{1 + \sin x}}{2} + t{g^2}x \\27)|\cos x| + \sin 3x = 0 \\28)2{\cos ^3}x + \cos 2x + \sin x = 0 \\29){\sin ^2}(x + \frac{\pi }{3}) - 6\sin (x + \frac{\pi }{3}) - 8 + {\cos ^2}5x = 0 \\30)\frac{{\sin 3x - \sin x}}{{\sqrt {1 - \cos 2x} }} = \sin 2x + \cos 2x,x \in (0,2\pi ) \\31)\sqrt {\cos 2x} + \sqrt {1 - \sin 2x} = 2\sqrt {\sin x - \cos x} \\32)\sin x - 2\sin 2x - \sin 3x = 2\sqrt 2 \\33)|\cot gx| = tgx + \frac{1}{{\sin x}} \\34)3\sin (x - \frac{\pi }{3}) + 4\sin (x + \frac{\pi }{6}) + 5\sin (5x + \frac{\pi }{6}) = 0 \\35)8\sin x = \frac{{\sqrt 3 }}{{\cos x}} + \frac{1}{{\sin x}} \\36)Tim{\rm{ }}a{\rm{ }}duong{\rm{ }}nho{\rm{ }}nhat{\rm{ }}thoa{\rm{ }}dieu{\rm{ }}kien: \\\cos [\pi ({a^2} + 2a - \frac{1}{2})] - \sin (\pi {a^2}) = 0 \\37)\frac{{{{\cos }^3}x - {{\sin }^3}x}}{{\sqrt {\cos x} + \sqrt {\sin x} }} = 2\cos 2x \\38)\sin xtg2x + \sqrt 3 (\sin x - \sqrt 3 tg2x) = 3\sqrt 3 \\Thoa{\rm{ }}man{\rm{ }}2{\rm{ }} + {\rm{ }}{\log _{\frac{1}{2}}}x \le 0 \\39)\sqrt {\cos x} + \cos x + {\cos ^2}x + \sin x = 1 \\40){\cos ^3}x\cos 3x + {\sin ^3}x\sin 3x = \frac{{\sqrt 2 }}{4}[/TEX]
Mong mọi người giúp đỡ.
Mình có 1 số bài lượng giác làm không ra, nay mình post lên đây để mọi người cùng tham khảo, làm va học tập, nhờ các bạn giúp đỡ:
[TEX]1) 3\cos x + 2|\sin x| = k;k = 2;k = 3\\2)\frac{{1 - \cos 4x}}{{2\sin 2x}} = \frac{{\sin 4x}}{{1 + \cos 4x}}[/TEX]
[TEX]\\ 3){\rm{ }}6tgx{\rm{ }} + {\rm{ }}a\cot g3x{\rm{ }} = {\rm{ }}tg2x.{\rm{ }}Giai{\rm{ }}khi{\rm{ }}a{\rm{ }} = {\rm{ }}0{\rm{ }}va{\rm{ }}a{\rm{ }} = {\rm{ }}5 \\4){\rm{ }}|\sin x{\rm{ }} - {\rm{ }}\cos x|{\rm{ }} + {\rm{ }}4\sin 2x{\rm{ }} = {\rm{ }}1 [/TEX]
[TEX]\\5){\rm{ }}t{g^2}x = \frac{{1 - {{\cos }^3}x}}{{1 - {{\sin }^3}x}} \\6){\rm{ }}t{g^2}x = \frac{{1 - \cos |x|}}{{1 - \sin |x|}} \\7){\rm{ }}\cos x{\rm{ }} + {\rm{ }}\frac{1}{{\cos x}} + \sin x + \frac{1}{{\sin x}} = \frac{{10}}{3} \\8){\rm{ }}3\cos x{\rm{ }} + {\rm{ }}4\sin x{\rm{ }} + {\rm{ }}\frac{6}{{3\cos x{\rm{ }} + {\rm{ }}4\sin x + 1}} = 6 [/TEX]
[TEX]\\9){\rm{ }}4{\cos ^2}x + 3t{g^2}x - 4\sqrt 3 \cos x + 2\sqrt 3 tgx + 4 = 0 \\10)3tg3x + \cot g2x = 2tgx + \frac{2}{{\sin 4x}} \\11)tg({120^0} + 3x) - tg({140^0} - x) = 2\sin ({80^0} + 2x) \\12)2(tgx - \sin x) + 3(\cot gx - \cos x) + 5 = 0 \\13){\sin ^3}x(1 + \cot gx) + {\cos ^3}x(1 + tgx) = 2\sqrt {\sin x\cos x} \\14)\frac{{{{\sin }^3}x + {{\cos }^3}x}}{{2\cos x - \sin x}} = \cos 2x \\15)\cos 2x - t{g^2}x = \frac{{{{\cos }^2}x - {{\cos }^3}x - 1}}{{{{\cos }^2}x}},x \in [1;70]rad \\16){\sin ^3}x + {\cos ^3}x = 2 - {\sin ^4}x \\17){\cos ^4}x - {\sin ^4}x = |\cos x| + |\sin x| \\18){\sin ^2}x + {\sin ^2}y + {\sin ^2}(x + y) = \frac{9}{4} \\19)2\sin (3x + \frac{\pi }{4}) = \sqrt {1 + 8\sin 2x{{\cos }^2}2x} \\20)tgx + t{g^2}x + t{g^3}x + \cot gx + \cot {g^2}x + \cot {g^3}x = 6 \\21)\cot g2x + \cot g3x + \frac{1}{{\sin x\sin 2x\sin 3x}} = 0 \\22){(tgx + \frac{1}{4}\cot gx)^n} = {\cos ^n}x + {\sin ^n}x,(n = 2,3,4...) \\23)\frac{{{{\sin }^{10}}x + {{\cos }^{10}}x}}{4} = \frac{{{{\sin }^6}x + {{\cos }^6}x}}{{4{{\cos }^2}2x + {{\sin }^2}2x}} \\24)\cos 2x - \cos 6x + 4(3\sin x - 4{\sin ^3}x + 1) = 0 [/TEX]
[TEX]\\25){\left( {{{\sin }^3}\frac{x}{2} + \frac{1}{{{{\sin }^3}\frac{x}{2}}}} \right)^2} + {\left( {{{\cos }^3}\frac{x}{2} + \frac{1}{{{{\cos }^3}\frac{x}{2}}}} \right)^2} = \frac{{81}}{4}{\cos ^2}4x \\26)\frac{{{{\sin }^4}\frac{x}{2} + {{\cos }^4}\frac{x}{2}}}{{1 - \sin x}} - t{g^2}x\sin x = \frac{{1 + \sin x}}{2} + t{g^2}x \\27)|\cos x| + \sin 3x = 0 \\28)2{\cos ^3}x + \cos 2x + \sin x = 0 \\29){\sin ^2}(x + \frac{\pi }{3}) - 6\sin (x + \frac{\pi }{3}) - 8 + {\cos ^2}5x = 0 \\30)\frac{{\sin 3x - \sin x}}{{\sqrt {1 - \cos 2x} }} = \sin 2x + \cos 2x,x \in (0,2\pi ) \\31)\sqrt {\cos 2x} + \sqrt {1 - \sin 2x} = 2\sqrt {\sin x - \cos x} \\32)\sin x - 2\sin 2x - \sin 3x = 2\sqrt 2 \\33)|\cot gx| = tgx + \frac{1}{{\sin x}} \\34)3\sin (x - \frac{\pi }{3}) + 4\sin (x + \frac{\pi }{6}) + 5\sin (5x + \frac{\pi }{6}) = 0 \\35)8\sin x = \frac{{\sqrt 3 }}{{\cos x}} + \frac{1}{{\sin x}} \\36)Tim{\rm{ }}a{\rm{ }}duong{\rm{ }}nho{\rm{ }}nhat{\rm{ }}thoa{\rm{ }}dieu{\rm{ }}kien: \\\cos [\pi ({a^2} + 2a - \frac{1}{2})] - \sin (\pi {a^2}) = 0 \\37)\frac{{{{\cos }^3}x - {{\sin }^3}x}}{{\sqrt {\cos x} + \sqrt {\sin x} }} = 2\cos 2x \\38)\sin xtg2x + \sqrt 3 (\sin x - \sqrt 3 tg2x) = 3\sqrt 3 \\Thoa{\rm{ }}man{\rm{ }}2{\rm{ }} + {\rm{ }}{\log _{\frac{1}{2}}}x \le 0 \\39)\sqrt {\cos x} + \cos x + {\cos ^2}x + \sin x = 1 \\40){\cos ^3}x\cos 3x + {\sin ^3}x\sin 3x = \frac{{\sqrt 2 }}{4}[/TEX]
Mong mọi người giúp đỡ.
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