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if cos x cos (pi/7)+sin x sin(pi/7)=-(sqrt 2)/2,then x can equal
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cos x cos (pi/7)+sin x sin(pi/7) this is cos(x - pi/7) so cos(x-pi/7) = -(sqrt2)/2 this means x-pi/7 = 3pi/4 or x-pi/7 = 5pi/4 now solve for x....
cos(x)cos(pi/7) + sin(x)sin(pi/7) = cos( x - pi/7) so \[\cos(x - \pi/7) = - \sqrt{2}/2\] the angles are in the 2nd (180 - x) and 3rd (180 + x) quadrants since cos is negative \[\cos^{-1}(\sqrt{2}/2) = \pi/4\] then 2rd quadrant \[(x - \pi/7) = 3\pi/4 \] \[x = 25\pi/28\] 3rd quadrant \[(x - \pi/7) = 5\pi/4\] \[x = 39\pi/28\]
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