# How do you solve 5\sin^2 (x)+8\sin x \cos x-3=0? I have

How do you solve $$\displaystyle{5}{{\sin}^{{2}}{\left({x}\right)}}+{8}{\sin{{x}}}{\cos{{x}}}-{3}={0}$$?
I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into $$\displaystyle{\sin{{2}}}{x}$$ but then I have $$\displaystyle{{\sin}^{{2}}{x}}$$ and $$\displaystyle{\sin{{2}}}{x}$$ together.

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alexandrebaud43
Hint:
$$\displaystyle{\frac{{{5}{{\sin}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}+{\frac{{{8}{\sin{{x}}}}}{{{\cos{{x}}}}}}-{\frac{{{3}{{\sin}^{{2}}{x}}+{3}{{\cos}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}={0}$$
$$\displaystyle{2}{{\tan}^{{2}}{x}}+{8}{\tan{{x}}}-{3}={0}$$
###### Not exactly what youâ€™re looking for?
Hattie Schaeffer
$$\displaystyle{2}{\left({5}{{\sin}^{{2}}{\left({x}\right)}}+{8}{\sin{{x}}}{\cos{{x}}}-{3}\right)}={5}{\left({1}-{\cos{{2}}}{x}\right)}+{8}{\sin{{2}}}{x}-{6}=-{5}{\cos{{2}}}{x}+{8}{\sin{{2}}}{x}-{1}={0}$$
This is a classical linear trigonometric equation
8s=5c+1
$$\displaystyle{64}{\left({1}-{c}^{{2}}\right)}={\left({5}{c}+{1}\right)}^{{2}}$$
$$\displaystyle{c}={\frac{{-{5}\pm{16}\sqrt{{{22}}}}}{{{89}}}},\ {s}={\frac{{{5}{c}+{1}}}{{{8}}}}$$
Vasquez

Rewrite it as $$5 \sin^2 x+8 \sin x \cos x-3(\sin^2 x+\cos^2 x)$$ ,which gives
$$2 \sin^2 x+8 \sin x \cos x-3 \cos^2 x$$ , a quadratic in $$\sin x$$
Using the quadratic formula, the roots of $$2u^2+8u-3$$ are $$-2-\sqrt{\frac{11}{2}}$$ and $$-2+\sqrt{\frac{11}{2}}$$ Hence this is equivalent to $$(u+2+\sqrt{\frac{11}{2}})(u+2-\sqrt{\frac{11}{2}})=0$$ so the original factorises as $$(\sin x+(2+\sqrt{\frac{11}{2}})\cos x)(\sin x+(2-\sqrt{\frac{11}{2}})\cos x)=0$$