How do you do this trig problem?
- Thread starter Faith54
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Is the above your answer? If so, what was the original exercise? Or if the above is the exercise, what are the instructions, what did you do, and what was your answer?Faith54 said:
Please reply with specifics. Thank you.
Eliz.
Hello, Faith54!
\(\displaystyle \;\;\)you played the \(\displaystyle K\spade\) instead of \(\displaystyle 5\club\)
\(\displaystyle \;\;\)and you transposed to \(\displaystyle G\) minor instead of \(\displaystyle B\flat\) major.
But seriously . . . we have: \(\displaystyle \,2\cdot\tan^22\theta \:= \:3\cdot\sec2\theta\)
Since \(\displaystyle \tan^2A\:=\:\sec^2A\,-\,1\), we have: \(\displaystyle \:2(\sec^22\theta\,-\,1)\;=\;2\cdot\sec2\theta\)
And we have the quadratic: \(\displaystyle \,2\cdot\sec^22\theta\,-\,3\cdot\sec2\theta\,-\,2\;=\;0\)
\(\displaystyle \;\;\)which factors: \(\displaystyle \
2\cdot\sec2\theta\,+\,1)(\sec2\theta\,-\,2)\;=\;0\)
And we have two equations to solve:
\(\displaystyle \;\;2\cdot\sec2\theta\,+\,1\:=\:0\;\;\Rightarrow\;\;\sec2\theta\,=\,-\frac{1}{2}\;\) . . . impossible! \(\displaystyle \;|\sec A|\,\geq\,1\)
\(\displaystyle \;\;\sec2\theta\,-\,2\:=\:0\;\;\Rightarrow\;\;\sec2\theta\,=\,2\;\;\Rightarrow\;\;2\theta\:=\:\pm\frac{\pi}{3}\,+\,2\pi n\;\;\Rightarrow\;\;\theta\:=\:\pm\frac{\pi}{6}\,+\,\pi n\)
It's hard to see your work from here, but I see two mistakes:
\(\displaystyle \;\;\)you played the \(\displaystyle K\spade\) instead of \(\displaystyle 5\club\)
\(\displaystyle \;\;\)and you transposed to \(\displaystyle G\) minor instead of \(\displaystyle B\flat\) major.
But seriously . . . we have: \(\displaystyle \,2\cdot\tan^22\theta \:= \:3\cdot\sec2\theta\)
Since \(\displaystyle \tan^2A\:=\:\sec^2A\,-\,1\), we have: \(\displaystyle \:2(\sec^22\theta\,-\,1)\;=\;2\cdot\sec2\theta\)
And we have the quadratic: \(\displaystyle \,2\cdot\sec^22\theta\,-\,3\cdot\sec2\theta\,-\,2\;=\;0\)
\(\displaystyle \;\;\)which factors: \(\displaystyle \
2\cdot\sec2\theta\,+\,1)(\sec2\theta\,-\,2)\;=\;0\)And we have two equations to solve:
\(\displaystyle \;\;2\cdot\sec2\theta\,+\,1\:=\:0\;\;\Rightarrow\;\;\sec2\theta\,=\,-\frac{1}{2}\;\) . . . impossible! \(\displaystyle \;|\sec A|\,\geq\,1\)
\(\displaystyle \;\;\sec2\theta\,-\,2\:=\:0\;\;\Rightarrow\;\;\sec2\theta\,=\,2\;\;\Rightarrow\;\;2\theta\:=\:\pm\frac{\pi}{3}\,+\,2\pi n\;\;\Rightarrow\;\;\theta\:=\:\pm\frac{\pi}{6}\,+\,\pi n\)