Solution for (cos2x)(csc^2x)=2cos2x & Sinx=cos2x?
- Thread starter Timcago
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Hello, Timcago!
A) \(\displaystyle \(\cos2x)(\csc^2x)\:=\:2\cdot\cos2x\;\) on the interval [0,2)
Then: \(\displaystyle \
\cos2x)(\csc^2x)\,-\,2\cdot\cos2x\;=\;0\)
Factor: \(\displaystyle \:\cos2x\left(\csc^2x\,-\,2\right)\;=\;0\)
And we have two equations to solve:
\(\displaystyle \;\;\cos2x\:=\:0\;\;\Rightarrow\;\;2x\,=\,\frac{\pi}{2},\;\frac{3\pi}{2}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{4},\;\frac{3\pi}{4}\)
\(\displaystyle \;\;\csc^2x\,=\,2\;\;\Rightarrow\;\;\csc x\,=\,\pm\sqrt{2}\;\;\Rightarrow\;\;\sin x\,=\,\pm\frac{1}{\sqrt{2}}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{4},\;\frac{3\pi}{4}\)
But \(\displaystyle \frac{3\pi}{4}\,>\,2\) . . . Therefore, the only solution is: \(\displaystyle \,x\,=\,\frac{\pi}{4}\)
B) \(\displaystyle \sin x \:=\:\cos2x\;\) ... all real solutions
We have: \(\displaystyle \,\sin x\:=\:1\,-\,2\sin^2x\;\;\Rightarrow\;\;2\sin^2x\,+\,
sin x \,- \,1\:=\:0\)
Factor: \(\displaystyle \,(2\sin x\,-\,1)(\sin x\,+\,1)\:=\:0\)
And we have two equations to solve:
\(\displaystyle \;\;2\sin x\,-\,1\:=\:0\;\;\Rightarrow\;\;\sin x\,=\,\frac{1}{2}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{6}\,+\,2\pi n,\;\frac{5\pi}{6}\,+\,2\pi n\)
\(\displaystyle \;\;\sin x\,+\,1\:=\:0\;\;\Rightarrow\;\;\sin x\,=\,-1\;\;\Rightarrow\;\;x\,=\,\frac{3\pi}{2}\,+\,2\pi n\)
A) \(\displaystyle \(\cos2x)(\csc^2x)\:=\:2\cdot\cos2x\;\) on the interval [0,2)
Then: \(\displaystyle \
\cos2x)(\csc^2x)\,-\,2\cdot\cos2x\;=\;0\)Factor: \(\displaystyle \:\cos2x\left(\csc^2x\,-\,2\right)\;=\;0\)
And we have two equations to solve:
\(\displaystyle \;\;\cos2x\:=\:0\;\;\Rightarrow\;\;2x\,=\,\frac{\pi}{2},\;\frac{3\pi}{2}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{4},\;\frac{3\pi}{4}\)
\(\displaystyle \;\;\csc^2x\,=\,2\;\;\Rightarrow\;\;\csc x\,=\,\pm\sqrt{2}\;\;\Rightarrow\;\;\sin x\,=\,\pm\frac{1}{\sqrt{2}}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{4},\;\frac{3\pi}{4}\)
But \(\displaystyle \frac{3\pi}{4}\,>\,2\) . . . Therefore, the only solution is: \(\displaystyle \,x\,=\,\frac{\pi}{4}\)
B) \(\displaystyle \sin x \:=\:\cos2x\;\) ... all real solutions
We have: \(\displaystyle \,\sin x\:=\:1\,-\,2\sin^2x\;\;\Rightarrow\;\;2\sin^2x\,+\,
sin x \,- \,1\:=\:0\)
Factor: \(\displaystyle \,(2\sin x\,-\,1)(\sin x\,+\,1)\:=\:0\)
And we have two equations to solve:
\(\displaystyle \;\;2\sin x\,-\,1\:=\:0\;\;\Rightarrow\;\;\sin x\,=\,\frac{1}{2}\;\;\Rightarrow\;\;x\,=\,\frac{\pi}{6}\,+\,2\pi n,\;\frac{5\pi}{6}\,+\,2\pi n\)
\(\displaystyle \;\;\sin x\,+\,1\:=\:0\;\;\Rightarrow\;\;\sin x\,=\,-1\;\;\Rightarrow\;\;x\,=\,\frac{3\pi}{2}\,+\,2\pi n\)
ohhhhh yea, i remember doing those now. I am doing the review for the final exam right now and i completely blanked when i saw that.
I had a typo. The interval is actually [0,2*PI)
Anyways on part A we have sinx=Plus or minus the root of 2 over 2.
Wouldnt that include PI/4, 3PI/4, 5PI/4, 7PI/4 since its both positive and negative root of 2 over 2?
I had a typo. The interval is actually [0,2*PI)
Anyways on part A we have sinx=Plus or minus the root of 2 over 2.
Wouldnt that include PI/4, 3PI/4, 5PI/4, 7PI/4 since its both positive and negative root of 2 over 2?