trig equation solution help

jislonica

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Jan 15, 2019
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can someone please assist with

[1+cosx]/sinx] + [sinx]/[1+cosx] = 4

given answer is sinx = 1/2 and i cannot get this... tk u
 

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can someone please assist with

[1+cosx]/sinx] + [sinx]/[1+cosx] = 4

given answer is sinx = 1/2 and i cannot get this... tk u

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?
 

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?
 
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