Paligutanhk

2022-01-03

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

alexandrebaud43

Hint:
$\frac{5{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}+\frac{8\mathrm{sin}x}{\mathrm{cos}x}-\frac{3{\mathrm{sin}}^{2}x+3{\mathrm{cos}}^{2}x}{{\mathrm{cos}}^{2}x}=0$
$2{\mathrm{tan}}^{2}x+8\mathrm{tan}x-3=0$

Hattie Schaeffer

$2\left(5{\mathrm{sin}}^{2}\left(x\right)+8\mathrm{sin}x\mathrm{cos}x-3\right)=5\left(1-\mathrm{cos}2x\right)+8\mathrm{sin}2x-6=-5\mathrm{cos}2x+8\mathrm{sin}2x-1=0$
This is a classical linear trigonometric equation
8s=5c+1
$64\left(1-{c}^{2}\right)={\left(5c+1\right)}^{2}$

Vasquez

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

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