cos (a-x) = cos x , find x

(h)[cos(a-x)] = (k)[cos x] , 90>a-x>0 , 90>x>0 ,x≠a find x cos (a – x) = (k/h) cos xcos a cos x + sin a sin x = (k/h)) cos xDivide both sides by cos xcos a + sin a sin x/cosx = (k/h) cos a + sin a tan x = (k/h) sin a tan x = (k/h) –cos a tan x =( k/h - cos a )/ sin a x = tan^( -1)[( k/h – cos a)/sin a ] (answer)
or x = arctan [( k/h – cos a)/sin a ]

Note:You have to know the formula for compound anglescos (A – B) = cos A cos B + sin A sin B If tan x = a, x = tan^(-1)(a) or arctan (a)arctan is the inverse trigonometric function