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\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
=>\(\dfrac{sin^2a+1}{cos^2a}+\dfrac{cos^2a+1}{sin^2a}=7\)
=>\(\dfrac{sin^4a+sin^2a+cos^4a+cos^2a}{sin^2a\cdot cos^2a}=7\)
=>\(sin^4a+cos^4a+1=7\cdot sin^2a\cdot cos^2a\)
=>\(\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a+1=7\cdot sin^2a\cdot cos^2a\)
=>\(2=9\cdot sin^2a\cdot cos^2a\)
=>\(8=9\cdot sin^22a\)
=>16=9(1-cos4a)
=>1-cos4a=16/9
=>cos4a=-7/9
1.
Kiểm tra lại đề bài, câu này phải là \(\dfrac{sinx+2cosx+3}{2sinx+cosx+3}\) mới đúng
2.a
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\dfrac{1}{cos^2x}=4tanx+6\)
\(\Leftrightarrow1+tan^2x=4tanx+6\)
\(\Leftrightarrow tan^2x-4tanx-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(5\right)+k\pi\end{matrix}\right.\)
2b.
Đặt \(x-\dfrac{\pi}{4}=t\Rightarrow x=t+\dfrac{\pi}{4}\)
\(sin^3t=\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow sin^3t=sint+cost\)
\(\Leftrightarrow sint\left(1-cos^2t\right)=sint+cost\)
\(\Leftrightarrow sint.cos^2t+cost=0\)
\(\Leftrightarrow cost\left(sint.cost+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\sin2t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\sin\left(2x-\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow...\)
270 độ<x<360 độ
=>sinx<0 và cosx>0
\(cos2x=\dfrac{2}{3}\)
=>\(2\cdot cos^2x-1=\dfrac{2}{3}\)
=>\(2\cdot cos^2x=\dfrac{5}{3}\)
=>\(cos^2x=\dfrac{5}{6}\)
mà cosx>0
nên \(cosx=\dfrac{\sqrt{30}}{6}\)
=>\(sinx=-\dfrac{\sqrt{6}}{6}\)
\(sin\left(x-\dfrac{pi}{6}\right)=sinx\cdot cos\left(\dfrac{pi}{6}\right)-cosx\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=-\dfrac{\sqrt{6}}{6}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{30}}{6}\cdot\dfrac{1}{2}=\dfrac{-3\sqrt{2}-\sqrt{30}}{12}\)
\(cos\left(x-\dfrac{pi}{6}\right)=cosx\cdot cos\left(\dfrac{pi}{6}\right)+sinx\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{\sqrt{30}}{6}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{-\sqrt{6}}{6}\cdot\dfrac{1}{2}=\dfrac{\sqrt{90}-\sqrt{6}}{12}\)