# SOLVING TRIGONOMETRIC EQUATIONS

\displaystyle\frac{{\pi}}{{2}},\frac{{{3}\pi}}{{2}};\frac{{{2}\pi}}{{3}},{\quad\text{and}\quad}\frac{{{4}\pi}}{{3}} Explanation:Rewrite the equation:f(x) = (sin 2x - cos 2x) - (sin x - cos x) ...

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How do you find the general solutions for \displaystyle{6}{\sin{{x}}}{\cos{{x}}}+{3}{\sin{{x}}}={2}{\cos{{x}}}+{1} ?
https://socratic.org/questions/how-do-you-find-the-general-solutions-for-6sin-x-cos-x-3sin-x-2cos-x-1
Noah G Mar 19, 2018 Put all terms to one side of the equation . \displaystyle{6}{\sin{{x}}}{\cos{{x}}}-{2}{\cos{{x}}}+{3}{\sin{{x}}}-{1}={0}\displaystyle{2}{\cos{{x}}}{\left({3}{\sin{{x}}}-{1}\right)}+{\left({3}{\sin{{x}}}-{1}\right)}={0} ...
We need to solve that \cos2x-\sin2x=\sqrt3(\cos2x-\sin2x)(\cos2x+\sin2x), which gives \cos2x-\sin2x=0, which is x=\frac{\pi}{8}+\frac{\pi}{2}k, k\in\mathbb Z or \cos2x+\sin2x=\frac{1}{\sqrt3}, ...

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Suppose a\sin x+b\cos x is a constant c. Plugging in x=0,\pi we get respectively a\cdot 0+b\cdot 1=c and a\cdot 0+b\cdot(-1)=c, i.e. b=c=-b. From there it follows that c=0, so 0 is the ...
Those are very general questions: what "can you do" and what "can't you do"...in terms of using trig identities: What you can do is use an identity to replace one expression with its strict ...

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The main formula (you either know it or you don't - meaning either it has been taught in class or your book, or if not then it's an unfair problem) is this: \sin A + \sin B = 2 \sin \frac {A+B} 2 \cos \frac {A-B} 2 ...
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