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To solve sinx-cosx = 0. Since the solution from giorgiana1976 and neela is not the same, one of you must be wrong?
x= pi/4 + k*pi, is not the same as
x= pi/4+k*pi or x= 3pi/4+k*pi
How is that?
To solve sinx-cosx = 0.
Since sin^2x+cos^2=1 is an identity, cosx = (1-sin^2x)^(1/2). So replacing cosx in the given equation we get:
sinx +(1-sin^2)^(1/2) = 0. Or
(1-sin^2x)^(1/2)=-sinx. squaring both sides,
1-sin^2x = sin^2x.Or
sin^2x = 1/2. Or
sinx= sqrt(1/2) or sinx = -sqrt(1/2) .
So x = pi/4 or 3pi/4 Or x= -pi/4 or 5pi/4 Or n*pi/4, where n is an integer.
Or x= 45 degree, 135deg,225 deg or 315deg . or n*45deg where n is an integer.
As we can see, the equation is linear, in sin x and cos x. The way of solving this kind of equation is to divide enire equation with "cos x".
(sin x/ cos x) - 1 = 0
Instead of the ratio "sin x/ cos x', we'll use the resulting function, tg x.
The obtained equation is an elementary equation.
tg x - 1 = 0
x= arctg 1 +k*pi
x= pi/4 + k*pi
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