How do you solve $3sinx=3-3cosx$ for $0<=x<=2pi$ ?

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Answer 1 · 2016-11-12T21:31:59

${0, pi/2}$

Alternative Approach:

$3sinx + 3cosx = 3$

$3(sinx + cosx) = 3$

$sinx + cosx = 1$

$(sinx+ cosx)^2 = 1^2$

$sin^2x+ 2sinxcosx + cos^2x = 1$

$1 + 2sinxcosx = 1$

$2sinxcosx= 0$

$(sinx)(cosx) = 0$

$x = 0, pi/2, pi, (3pi)/2, 2pi$

Checking in the original equation, you will find that only $x = pi/2$ and $x= 0$ are proper solutions.

Hopefully this helps!

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