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[momaru12]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
Chắc suất Đại học top - Giữ chỗ ngay!!

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[TEX]1.\sqrt{3 - cosx}- \sqrt{cosx +1} = 2\\2. 2sin (3x + \frac{pi}{4}) = \sqrt{1 + 8sin2x cos2x}\\3.4\sqrt{3}sinx . cosx . cos2x=sin 8x\\4. sin^2x + sin^22x + sin^23x = \frac{3}{2}\\5. sin^8x + cos^8x = 2(sin^10x + cos^10x) +\frac{5}{4}cos2x\\6. \sqrt{3} sin2x - 2cos^2x = 2\sqrt{2+ 2cos2x}\\7. \frac{3(sinx + tgx)}{tgx-sinx}-2cosx = 0[/TEX]

 

[vurain]

[tex](3) <=> 2\sqrt{3} sin 2x.cos2x=sin8x [/tex]
[tex]<=> \sqrt{3}sin4x=sin8x [/tex]
[tex]<=> \sqrt{3}sin4x=2sin4xcos4x [/tex]
ban lam típ nha

 

[vurain]

[tex](4) <=> 1-cos2x+1-cos4x+1-cos6x=3<=>cos6x +cos2x+cos4x=0 [/tex]


[tex]<=> 2cos4xcos2x+cos4x=0 [/tex]

 

[duoisam117]

1
5. sin^8x + cos^8x = 2(sin^10x + cos^10x) + 5/4 cos2x

[TEX]sin^8x + cos^8x = 2(sin^{10}x + cos^{10}x) + \frac{5}{4} cos2x[/TEX]

[TEX]\Leftrightarrow (sin^8x-2sin^{10}x ) + (cos^8x-2 cos^{10}x) = \frac{5}{4} cos2x[/TEX]

[TEX]\Leftrightarrow sin^8x (1-2sin^2x ) - cos^8x (-1+2 cos^2x) = \frac{5}{4} cos2x[/TEX]

[TEX]\Leftrightarrow sin^8x .cos2x - cos^8x .cos2x = \frac{5}{4} cos2x[/TEX]

[TEX]\Leftrightarrow 4cos2x .( sin^8x- cos^8x) = 5 cos2x[/TEX]

[TEX]\Leftrightarrow cos2x.[4(sin^8x-cos^8x)-5]=0[/TEX]

[TEX]\Leftrightarrow cos2x.[4(sin^4x-cos^4x). (sin^4x+cos^4x)-5]=0[/TEX]

[TEX]\Leftrightarrow cos2x.[4(sin^2x-cos^2x). (1-\frac{1}{2}sin^22x)-5]=0[/TEX]

Tới đây rùi làm sao zạ Tui pí rùi

 

[rua_it]

[TEX]sin^8x + cos^8x = 2(sin^{10}x + cos^{10}x) + \frac{5}{4} cos2x[/TEX]

[TEX]\Leftrightarrow (sin^8x-2sin^{10}x ) + (cos^8x-2 cos^{10}x) = \frac{5}{4} cos2x[/TEX]

[TEX]\Leftrightarrow sin^8x (1-2sin^2x ) - cos^8x (-1+2 cos^2x) = \frac{5}{4} cos2x[/TEX]

[TEX]\Leftrightarrow sin^8x .cos2x - cos^8x .cos2x = \frac{5}{4} cos2x[/TEX]

[TEX]\Leftrightarrow 4cos2x .( sin^8x- cos^8x) = 5 cos2x[/TEX]

[TEX]\Leftrightarrow cos2x.[4(sin^8x-cos^8x)-5]=0[/TEX]

[TEX]\Leftrightarrow cos2x.[4(sin^4x-cos^4x). (sin^4x+cos^4x)-5]=0[/TEX]

[TEX]\Leftrightarrow cos2x.[4(sin^2x-cos^2x). (1-\frac{1}{2}sin^22x)-5]=0[/TEX]

Tới đây rùi làm sao zạ Tui pí rùi

[tex]\mathrm{\blue{sin^8x + cos^8x = 2(sin^{10}x + cos^{10}x) + \frac{5}{4} cos2x[/TEX]

[TEX]\Leftrightarrow sin^8x .cos2x - cos^8x .cos2x = \frac{5}{4} cos2x[/TEX]

[tex]\Leftrightarrow cos2x.(sin^8x-cos^8x-\frac{5}{4})=0[/tex]

[tex]\mathrm{\Leftrightarrow cos2x=0 \Leftrightarrow x=\frac{\pi}{4}+\frac{k.\pi}{2} ( k \in\ Z) \ Do \ cos^8x+\frac{5}{4} > sin^8x[/tex]

 

[quyenuy0241]

[TEX]1.\sqrt{3 - cosx}- \sqrt{cosx +1} = 2[/tex]

[tex]cosx \ge -1 [/tex]

[tex]\sqrt{cosx+1} \ge 0 [/tex]

[tex]\sqrt{3-cosx} \le 2 [/tex]

Suy ra [tex]VT \le Vp [/tex]

Dấu = xảy ra: [tex]cosx=-1 [/tex]

 

[quyenuy0241]

[tex]\sqrt{3} sin2x - 2cos^2x = 2\sqrt{2+ 2cos2x}[/tex]

[tex]\sqrt{3}sin2x-2cos^2x=4cosx [/tex]

[tex]cosx(\sqrt{3}sinx-cosx-2)=0 [/tex]

[tex]\left[cosx=0 \\ sin(x-\frac{\pi}{6})=0[/tex]

[tex]7. \frac{3(sinx + tgx)}{tgx-sinx}-2cosx = 0[/tex]

DKXD:.....

[tex] \frac{3(sinx + tgx)}{tgx-sinx}=2cosx [/tex]

[tex]\Leftrightarrow 3sinx+3tgx=2sinx-sin2x \Leftrightarrow sinx+sin2x+3tgx=0\\\Leftrightarrow sinxcosx+2sinxcos^2x+3sinx=0[/tex]

[tex]\left{sinx=0(L) \\ cosx+2cos^2x+3=0(Vo.n_o) [/tex]8-X