How do you solve $cos x - 4 = sin x - 4$ $cosx-4=sinx-4$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2016-09-23T11:23:47

$x = \frac{π}{4}, x = \frac{5 π}{4}$ $x= pi/4, x=(5pi)/4$ for $0 \leq x \leq 2 π$ $0<= x <= 2pi$

$cosx-4=sinx-4 or cosx= sinx or tanx=1 ; tan(pi/4)=1 ; tan(pi+pi/4)=1 or tan((5pi)/4)=1 :. x= pi/4, x=(5pi)/4$ [Ans]

How do you solve cosx-4=sinx-4cosx−4=sinx−4cosx-4=sinx-4 for 0<=x<=2pi0≤x≤2π0<=x<=2pi?