Solve: cos(x)(cosx+1)=0
is it cos(x+1)?
or (cos x)+1
its (cosx+1)
like...i...said...
either cos(x) is 0 or cos(x+1) is 0
Set each to 0 and solve \[\cos(x) = 0\] \[\cos(x) + 1 = 0 or \cos(x) = -1\] cos(x) is 0 when x is pi/2 3pi/2, 5pi/2 ect so the answer would be \[x = \frac{\pi}{2} +\pi*n\] and cos(x) is -1 when x is pi, 3pi, 5pi,ect so the answer to that would be \[x = \pi + 2n*\pi\]
oh, i should mention, n is a integer value
The avalaible answers are A. x = PI/3+-2PIn, x = 3PI/4+-2PIn B. x = Pi/2+-2PIn, x = 3PI/2P-2PIn, x =Pi+-2PIn C. x =+-PIn, x = Pi/2+-2PIn D. x = Pi/2+-PIn, x = 3Pi/2+-2Pin
D
Wrong again but hey..tbh im not really surprized...everyone has given me wrong answers for the last 6 questions...why should i expect dif
wait hang on..
too late moving on thnx anyways tho
It was b ._.
ohh, i see what they did in B. They did it weird.
yeah well..im starting to get pissed ...one question wrong occasionally im fine with....but the last 6 ..well now 7 have gotten wrong
in.a.row. .....
Life is tough
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