Question

5sin^2(x)-14sin(x)cos(x)-3cos^2(x)=2 Показать полное решение

Answer 1

$5sin ^{2} x-14sinxcosx-3cos ^{2} x=2$
$5sin ^{2} x-14sinxcosx-3cos ^{2} x-2(sin ^{2}x+cos ^{2}x) =0$
$5sin ^{2} x-14sinxcosx-3cos ^{2} x-2sin ^{2}x-2cos ^{2}x =0$
$3sin ^{2} x-14sinxcosx-5cos ^{2} x =0 |: cos ^{2} x,$  $(cos ^{2} x \neq 0)$

$3 \frac{sin ^{2}x }{cos ^{2}x } -14 \frac{sinx}{cosx} -5=0$

$3tg ^{2} x-14tgx-5=0$
$tgx=y$
$3y ^{2} -14y-5=0$
D = (-14)² - 4· 3·(-5) = 196 + 60 = 256

$\sqrt{D}=16$

$y_{1} = \frac{14+16}{6} = \frac{30}{6}=5$

$y_{2} = \frac{14-16}{6} = \frac{-2}{6} =- \frac{1}{3}$

$\left \{ {{tgx=5} \atop {tgx=- \frac{1}{3} }} \right.$

$x=arctg5+ \pi n,$ n∈Z
$x=-arctg \frac{1}{3} + \pi m,$ m∈Z

Answer 2

I can help you with your math problem. Here is the detailed solution:

The given equation is $$5\sin^2(x)-14\sin(x)\cos(x)-3\cos^2(x)=2$$

We can use the identity $$\sin^2(x)+\cos^2(x)=1$$ to rewrite the equation as $$5(1-\cos^2(x))-14\sin(x)\cos(x)-3\cos^2(x)=2$$

Simplifying, we get $$-8\cos^2(x)-14\sin(x)\cos(x)+3=2$$

Dividing by -2, we get $$4\cos^2(x)+7\sin(x)\cos(x)-\frac{3}{2}=0$$

Let $$u=\cos(x)$$, then the equation becomes $$4u^2+7u-\frac{3}{2}=0$$

Using the quadratic formula, we get $$u=\frac{-7\pm\sqrt{49+24}}{8}$$

Simplifying, we get $$u=-\frac{3}{4} \text{ or } u=\frac{1}{2}$$

Substituting back $$u=\cos(x)$$, we get $$\cos(x)=-\frac{3}{4} \text{ or } \cos(x)=\frac{1}{2}$$

Using the inverse cosine function, we get $$x=\cos^{-1}(-\frac{3}{4}) \text{ or } x=\cos^{-1}(\frac{1}{2})$$

Using a calculator, we get $$x\approx 2.42 \text{ or } x\approx 1.05$$

However, these are not the only solutions, since the cosine function is periodic. We can find the general solutions by adding or subtracting multiples of $$2\pi$$ to the above values. Therefore, the final solutions are $$x=2.42+2k\pi \text{ or } x=1.05+2k\pi$$ where $$k$$ is any integer.5sin^2(x)-14sin(x)cos(x)-3cos^2(x)=2+Показать+полное+решение