# Solve: 3\sin 2x+4 \cos 2x− 2\cos x+6\sin x-6=0 My Try 6 \sin

Solve: $$\displaystyle{3}{\sin{{2}}}{x}+{4}{\cos{{2}}}{x}−{2}{\cos{{x}}}+{6}{\sin{{x}}}-{6}={0}$$
My Try
$$\displaystyle{6}{\sin{{x}}}{\cos{{x}}}+{4}{\left({{\cos}^{{2}}{x}}−{{\sin}^{{2}}{x}}\right)}−{2}{\cos{{x}}}+{6}{\sin{{x}}}−{6}={0}$$

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Timothy Wolff
$$\displaystyle{6}{\sin{{x}}}{\cos{{x}}}+{4}{\left({{\cos}^{{2}}{x}}−{{\sin}^{{2}}{x}}\right)}−{2}{\cos{{x}}}+{6}{\sin{{x}}}−{6}={0}$$
$$\displaystyle{6}{\sin{{x}}}{\cos{{x}}}−{9}{{\sin}^{{2}}{x}}−{{\cos}^{{2}}{x}}−{2}{\cos{{x}}}+{6}{\sin{{x}}}−{1}={0}$$
$$\displaystyle-{\left({3}{\sin{{x}}}−{\cos{{x}}}\right)}^{{2}}+{2}{\left({3}{\sin{{x}}}−{\cos{{x}}}\right)}−{1}={0}$$
$$\displaystyle{\left({3}{\sin{{x}}}−{\cos{{x}}}\right)}^{{2}}−{2}{\left({3}{\sin{{x}}}−{\cos{{x}}}\right)}+{1}={0}$$
###### Not exactly what you’re looking for?
Philip Williams
x=2y we get after some manipulation
$$\displaystyle-{4}{{\cos}^{{4}}{y}}−{36}{{\cos}^{{2}}{y}}{{\sin}^{{2}}{y}}+{24}{{\cos}^{{3}}{y}}{\sin{{y}}}={0}$$
$$\displaystyle-{4}{{\cos}^{{2}}{y}}{\left({\cos{{y}}}−{3}{\sin{{y}}}\right)}^{{2}}={0}$$
Vasquez

Abbreviating $$\sin x$$ and $$\cos x$$ to s and c, and writing $$\sin 2x=2sc$$ and $$\cos 2x=2c^2-1$$, the equation becomes
$$6sc+4(2c^2-1)-2c+6s-6=0$$
Factoring out a 2 and grouping terms, we rewrite this as
$$3s(c+1)+4c^2-c-5=0$$
But $$4c^2-c-5=(4c-5)(c+1)$$, so we have two solutions in s and c:
c=-1 and 3s+4c=5
Reverting to $$\sin x$$ and $$\cos x$$ and recognizing the Pythagorean triple $$3^2+4^2=5^2$$, these become
$$\cos x=-1$$ and $$\sin(x+\theta)=1$$
where $$\sin \theta=\frac 45$$ and $$\cos \theta=\frac 35$$, e.g. $$\theta=\arcsin(\frac 45)$$. So the solution set is
$$\{\pi+2n\pi | n\in \mathbb{Z}\} \cup \{ \frac{\pi}{2}−\arcsin(\frac 45)+2n\pi | n \in \mathbb{Z}\}$$