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Mathematics 12 Online
OpenStudy (anonymous):

(a) Solve for all values of cos x to the nearest tenth. 4 cos2 x = cos x + 2 (b) How many values of x between 180 degrees and 360 degrees satisfy the equation 4 cos2 x = cos x + 2?

OpenStudy (freckles):

maybe this identity could be useful cos(2x)=cos^2(x)-sin^2(x) =cos^2(x)-(1-cos^2(x)) =2cos^2(x)-1

OpenStudy (anonymous):

But how am I suppose to use this identity to find the values of x ??

OpenStudy (freckles):

you will have a quadtratic equation in terms of cos

OpenStudy (freckles):

\[4 \cos(2x)=\cos(x)+2 \\ 4(2 \cos^2(x)-1)=\cos(x)+2 \\8\cos^2(x)-4=\cos(x)+2 \\ 8\cos^2(x)-\cos(x)-6=0\] use the quadratic formula to solve for cos(x)

OpenStudy (freckles):

then you can solve for x

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