lượng giác 11 tiếp ạ

[connguoivietnam]

1)
[TEX]4cos^2x+3tan^2x-4\sqrt{3}cosx+2\sqrt{3}tanx+4=0[/TEX]

[TEX]4cos^2x-4\sqrt{3}cosx+3+3tan^2x-2\sqrt{3}tanx+1=0[/TEX]

[TEX](2cosx-\sqrt{3})^2+(\sqrt{3}tanx-1)^2=0[/TEX]

dấu bằng xảy ra khi

[TEX]2cosx-\sqrt{3}=0[/TEX]

[TEX]\sqrt{3}tanx-1=0[/TEX]

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3)
[TEX]sinx+\sqrt{1+cos^2x}+sinx\sqrt{+cos^2x}=3[/TEX]

đặt [TEX]sinx+\sqrt{1+cos^2x}=t (-2 \leq t \leq 2)[/TEX]

[TEX]sinx\sqrt{1+cos^2x}=\frac{t^2-2}{2}[/TEX]

[TEX]\Rightarrow t+\frac{t^2-2}{2}=3[/TEX]

[TEX]2t+t^2-2=6[/TEX]

[TEX]t^2+2t-8=0[/TEX]

[TEX]t=2(T/M)[/TEX]

[TEX]sinx+\sqrt{1+cos^2x}=2[/TEX]

 

[giaosu_fanting_thientai]

Tiếp nhé:
2/ [TEX]8cos4xcos^2 2x + 1 + \sqrt{1-cos3x} = 0[/TEX]

[/TEX][/B]

[TEX]<=>8(2cos^22x-1)cos^22x+1+\sqrt{1-cos3x}=0[/TEX]
[TEX]<=>16cos^42x-8cos^22x+1+\sqrt{1-cos3x}=0[/TEX]
[TEX]<=> (4cos^22x-1)^2+\sqrt{1-cos3x}=0[/TEX]
<=>[TEX] 4cos^22x-1=0[/TEX] Và [TEX]1-cos3x=0[/TEX]