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

Paligutanhk 2022-01-03 Answered
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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Expert Answer

alexandrebaud43
Answered 2022-01-04 Author has 127 answers
Hint:
\(\displaystyle{\frac{{{5}{{\sin}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}+{\frac{{{8}{\sin{{x}}}}}{{{\cos{{x}}}}}}-{\frac{{{3}{{\sin}^{{2}}{x}}+{3}{{\cos}^{{2}}{x}}}}{{{{\cos}^{{2}}{x}}}}}={0}\)
\(\displaystyle{2}{{\tan}^{{2}}{x}}+{8}{\tan{{x}}}-{3}={0}\)
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Hattie Schaeffer
Answered 2022-01-05 Author has 3318 answers
\(\displaystyle{2}{\left({5}{{\sin}^{{2}}{\left({x}\right)}}+{8}{\sin{{x}}}{\cos{{x}}}-{3}\right)}={5}{\left({1}-{\cos{{2}}}{x}\right)}+{8}{\sin{{2}}}{x}-{6}=-{5}{\cos{{2}}}{x}+{8}{\sin{{2}}}{x}-{1}={0}\)
This is a classical linear trigonometric equation
8s=5c+1
\(\displaystyle{64}{\left({1}-{c}^{{2}}\right)}={\left({5}{c}+{1}\right)}^{{2}}\)
\(\displaystyle{c}={\frac{{-{5}\pm{16}\sqrt{{{22}}}}}{{{89}}}},\ {s}={\frac{{{5}{c}+{1}}}{{{8}}}}\)
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Vasquez
Answered 2022-01-08 Author has 9499 answers

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 \((\sin x+(2+\sqrt{\frac{11}{2}})\cos x)(\sin x+(2-\sqrt{\frac{11}{2}})\cos x)=0\)

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