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Solve the following equation cot x-(square root of 5)cosec x=2 Find the maximum and minimum for cos x-2sinx+3 .
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cot x - sqrt5 csc x =2 (cos x - sqrt5)/sin x = 2 cos x - sqrt5 = 2sin x square both sides cos^2 -2sqrt5*cos x +5 = 4sin^2 = 4(1-cos^2) 5cos^2 -2sqrt5*cos x +1 = 0 (sqrt5cos(x) -1)^2 = 0 sqrt5*cos x = 1 cos x = 1/sqrt5 x = cos^-1 (1/sqrt5) = 63.43 degrees
oops, 63.43 degrees doesn't work the other possible case is 360 - 63.43 = 296.57 degrees this way cos still equals 1/sqrt5
to find max/min, differentiate and set equal to zero f'(x) = -sin x -2cos x = 0 -2cos x = sin x tan x = -2 x = tan^-1 (-2) x = 116.57 and 296.57 min = 116.57 max = 296.57
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