trig equation solution help
- Thread starter jislonica
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You were SO CLOSE!
\(\displaystyle \dfrac{2+2\cos(x)}{\sin(x)+\sin(x)\cos(x)} = \dfrac{2\cdot(1+\cos(x))}{\sin(x)\cdot(1+\cos(x))}\) -- Now what?
- Joined
- Apr 12, 2005
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Why the "?"? Look for errors. If you find none, smile and move on to the next problem. This builds confidence.
Don't be in a rush to EXPAND or to SIMPLIFY or to COMBINE. Sometimes, you want things FACTORED or SEPARATED.
Experience will help.\(\displaystyle \dfrac{1 + cos(x)}{sin(x)} + \dfrac{sin(x)}{1 + cos(x)} = 4 \implies\)
\(\displaystyle \left (\dfrac{1 + cos(x)}{sin(x)} * \dfrac{1 + cos(x)}{1 + cos(x)} \right ) + \left ( \dfrac{sin(x)}{1 + cos(x)} * \dfrac{sin(x)}{sin(x)} \right ) = 4 \implies\)
\(\displaystyle \dfrac{1 + 2cos(x) + cos^2(x)}{sin(x) * \{1 + cos(x)\}} + \dfrac{sin^2(x)}{sin(x) * \{1 + cos(x)\}} = 4 \implies\)
\(\displaystyle 4 = \dfrac{1 + 2cos(x) + \{cos^2(x) +sin^2(x)\} }{sin(x) * \{1 + cos(x)\}} = \dfrac{1 + 2cos(x) + 1}{sin(x) * \{1 + cos(x)\}}= \dfrac{2 + 2cos(x)}{sin(x) * \{1 + cos(x)\}} \implies\)
\(\displaystyle 4 = \dfrac{2\{1 + cos(x)\}}{sin(x) * \{1 + cos(x)\}} = \dfrac{2}{sin(x)} \implies\)
\(\displaystyle sin(x) = \dfrac{2}{4} = \dfrac{1}{2}.\)
Well done, but are you supposed to find x?
\(\displaystyle \left (\dfrac{1 + cos(x)}{sin(x)} * \dfrac{1 + cos(x)}{1 + cos(x)} \right ) + \left ( \dfrac{sin(x)}{1 + cos(x)} * \dfrac{sin(x)}{sin(x)} \right ) = 4 \implies\)
\(\displaystyle \dfrac{1 + 2cos(x) + cos^2(x)}{sin(x) * \{1 + cos(x)\}} + \dfrac{sin^2(x)}{sin(x) * \{1 + cos(x)\}} = 4 \implies\)
\(\displaystyle 4 = \dfrac{1 + 2cos(x) + \{cos^2(x) +sin^2(x)\} }{sin(x) * \{1 + cos(x)\}} = \dfrac{1 + 2cos(x) + 1}{sin(x) * \{1 + cos(x)\}}= \dfrac{2 + 2cos(x)}{sin(x) * \{1 + cos(x)\}} \implies\)
\(\displaystyle 4 = \dfrac{2\{1 + cos(x)\}}{sin(x) * \{1 + cos(x)\}} = \dfrac{2}{sin(x)} \implies\)
\(\displaystyle sin(x) = \dfrac{2}{4} = \dfrac{1}{2}.\)
Well done, but are you supposed to find x?