anonymous
  • anonymous
tan3x=-1 , solve for all solutions
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Mertsj
  • Mertsj
\[3x=\frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}, \frac{17\pi}{4}, \frac{21\pi}{4}\]
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=solve+tan%283+x%29%3D%3D-1
Mertsj
  • Mertsj
\[x=\frac{\pi}{12}, \frac{5\pi}{12}, \frac{9\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{21\pi}{12}+2\pi n\]

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anonymous
  • anonymous
would the answers be the same if the q was solve for cot(3x)=-1
Mertsj
  • Mertsj
Oh boy. I didn't see the negative sign. The answers for 3x should be in quadrants 2 and 4 instead of 1 and 3. So replace the numerators with these numbers:
Mertsj
  • Mertsj
\[3\pi, 7\pi,11\pi, 15\pi, 19\pi, 23\pi\]
Mertsj
  • Mertsj
And yes, since the cot is the reciprocal of the tangent, and since the reciprocal of -1 is -1, the answers would be the same.
anonymous
  • anonymous
can u help me with another one
Mertsj
  • Mertsj
Post it.
anonymous
  • anonymous
kay here it is csc^2(2x)-csc(2x)-2=0
Mertsj
  • Mertsj
\[\csc ^2(2x)-\csc (2x)-2=0\]
Mertsj
  • Mertsj
\[(\csc (2x)-2)(\csc (2x)+1)=0\]
Mertsj
  • Mertsj
\[\csc (2x)=2 , \csc (2x)=-1\]
Mertsj
  • Mertsj
\[2x=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, \frac{17\pi}{6}\]
Mertsj
  • Mertsj
\[x=\frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12} + 2\pi n\]
Mertsj
  • Mertsj
\[\csc (2x)=-1\]
anonymous
  • anonymous
oh and can u help me with a true and false question for this equeation: 3tan^2(2x)-1=0
anonymous
  • anonymous
I already got the question above lets move on
anonymous
  • anonymous
This equation has 8 solutions on the domain . is one of the solutions of this equation. The general solution of this equation is . is one of the solutions of this equation which option is not true
Mertsj
  • Mertsj
\[2x=\frac{3\pi}{2}, \frac{7\pi }{2}, x=\frac{3\pi}{4}, \frac{7\pi}{4} + 2\pi n\]
anonymous
  • anonymous
its missing some info
Mertsj
  • Mertsj
So how am I supposed to answer it?
anonymous
  • anonymous
so for the first option it was supposed to say in the domain of 0 to 2pi
anonymous
  • anonymous
the second option was supposed to say 13pi/12 is one of the solutions
Mertsj
  • Mertsj
8 solutions.
anonymous
  • anonymous
so the eight solutions one is the false one
anonymous
  • anonymous
one last question for you
Mertsj
  • Mertsj
I cannot tell which one is the false one because the answers are not posted.

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