Bài 172:
[TEX]\Large 2sin^{2}(\frac{\pi}{2}.cos^{2}x)=1-cos(\pi.sin2x)[/TEX]
[TEX]PT \Leftrightarrow 2sin^2( \frac{ \pi}{2}.cos^2x)=2sin^2( \frac{ \pi}{2}.sin2x)[/TEX]
[TEX]\Leftrightarrow \left[\begin{sin( \frac{ \pi}{2}cos^2x)=sin( \frac{ \pi}{2}sin2x)(1)}\\{sin( \frac{ \pi}{2}cos^2x)=-sin( \frac{ \pi}{2}sin2x)(2) }[/TEX]
Giải (1):Ta có:
[TEX](1) \Leftrightarrow \left[\begin{ \frac{ \pi}{2}cos^2x= \frac{ \pi}{2}sin2x+k2 \pi}\\{ \frac{ \pi}{2}cos^2x= \pi- \frac{ \pi}{2}sin2x+k2 \pi }[/TEX]
[TEX] \frac{ \pi}{2}cos^2x= \frac{ \pi}{2}sin2x+k2 \pi \Leftrightarrow sin2x-cos^2x+4k=0[/TEX]
[TEX]\Leftrightarrow 2sin2x-cos2x=1-4k (3)[/TEX]
(3) có nghiệm khi [TEX](1-4k)^2 \leq 2^2+1 \Leftrightarrow \frac{1- \sqrt{5}}{4} \leq k \leq \frac{1+ \sqrt{5}}{4} \Rightarrow k=0[/TEX]
[TEX]\Rightarrow 2sin2x-cos2x=1 \Rightarrow Ok[/TEX]
Tương tự với các TH còn lại
Bài 173:
[TEX]\Large 8^{sin^{2}x}+8^{cos^{2}x}=10+cos2y[/TEX]
[TEX]VP=10+cos2x \geq 9[/TEX]
Đặt [TEX]t=8^{sin^2x}, 1 \leq t \leq 8[/TEX]
[TEX]\Rightarrow VT=f(t)=t+ \frac{8}{t} \Rightarrow f'(t)=1- \frac{8}{t^2}=0 \Leftrightarrow t=2 \sqrt{2}[/TEX]
Lập BBT cho hàm f(t) ta suy ra:
[TEX]Maxf(t)=f(1)=f(8)=9 \Rightarrow VT \leq 9[/TEX]
Do vậy: [TEX]PT \Leftrightarrow \left{\begin{sinx=0,hoac:sin^2x=1}\\{cos2y=-1}[/TEX]
Bài 175:
[TEX]\Large \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[/TEX]
[TEX]VT=sin^6{ \frac{x}{2}}+cos^6{ \frac{x}{2}}+ \frac{1}{sin^6{ \frac{x}{2}}}+ \frac{1}{cos^6{ \frac{x}{2}}}+4[/TEX]
Đặt [TEX]a=sin^2{ \frac{x}{2}},b=cos^2{ \frac{x}{2}} \Rightarrow a+b=1 \Rightarrow ab \leq \frac{1}{4}[/TEX]
[TEX]VT=4+a^3+b^3+ \frac{1}{a^3}+ \frac{1}{b^3} \geq 4+2.( \frac{a+b}{2})^3+ \frac{2}{ \sqrt{a^3b^3}} \geq 4+ \frac{1}{4}+16= \frac{81}{4}[/TEX]
[TEX]\Rightarrow VT \geq \frac{81}{4} \geq \frac{81}{4}cos^24x=VP[/TEX]
Do vậy:
[TEX]PT \Leftrightarrow \left{\begin{sin^2{ \frac{x}{2}}=cos^2{ \frac{x}{2}}}\\{cos^42x=1}[/TEX]