Question

21sin^2 x+sin2x−15cos^2 x=2

Answer 1

Ответ:

$21 { \sin }^{2} x + \sin(2x) - 15 { \cos }^{2} x = 2 \\ 21 { \sin }^{2} x + 2 \sin(x) \cos(x) - 15 { \cos }^{2} x = 2 { \sin}^{2} x + 2 { \cos }^{2} x \\ 19 { \sin}^{2} x + 2\sin(x) \sin(x) - 17 { \cos}^{2} x = 0$

разделим на cos^2x, не равный 0.

$19 {tg}^{2} (x) + 2tg(x) - 17 = 0 \\ \\ tg(x) = t \\ 19 {t}^{2} + 2t - 17 = 0 \\ d = 4 + 1292 = 1296 \\ t1 = \frac{ - 2 + 36}{19 \times 2} = \frac{34}{2 \times 19} = \frac{17}{19} \\ t2 = \frac{ - 38}{2 \times 19} = - 1 \\ \\ tg(x) = \frac{17}{19} \\ x1 = arctg( \frac{17}{19} ) + \pi \: n \\ \\ tg(x) = - 1 \\ x2 = - \frac{\pi}{4} + \pi \: n$

n принадлежит Z.

Answer 2

To solve the equation $21\sin^2 x + \sin 2x - 15\cos^2 x = 2$, you can use trigonometric identities to simplify it and then find the values of $x$ that satisfy the equation.

Start by using the double angle identity for sine, which states that $\sin 2x = 2\sin x \cos x$. Also, you can use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to replace $\cos^2 x$ with $1 - \sin^2 x$. The equation becomes:

$21\sin^2 x + 2\sin x \cos x - 15(1 - \sin^2 x) = 2$

Now, simplify further:

$21\sin^2 x + 2\sin x \cos x - 15 + 15\sin^2 x = 2$

Combine like terms:

$36\sin^2 x + 2\sin x \cos x - 15 = 2$

Rearrange the equation:

$36\sin^2 x + 2\sin x \cos x - 17 = 0$

Now, you have a quadratic equation in terms of $\sin x$. You can solve this equation using the quadratic formula:

$\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 36$, $b = 2\cos x$, and $c = -17$.

$\sin x = \frac{-2\cos x \pm \sqrt{(2\cos x)^2 - 4(36)(-17)}}{2(36)}$

Simplify further:

$\sin x = \frac{-\cos x \pm \sqrt{4\cos^2 x + 4(36)(17)}}{72}$

$\sin x = \frac{-\cos x \pm \sqrt{4(\cos^2 x + 36\cdot 17)}}{72}$

$\sin x = \frac{-\cos x \pm \sqrt{4\cos^2 x + 4\cdot 36\cdot 17}}{72}$

$\sin x = \frac{-\cos x \pm \sqrt{4(\cos^2 x + 36\cdot 17)}}{72}$

$\sin x = \frac{-\cos x \pm \sqrt{4(\cos^2 x + 612)}}{72}$

Now, you can solve for $\sin x$ in terms of $\cos x$ and then find the corresponding values of $x$. Keep in mind that $\sin x$ and $\cos x$ are related by the Pythagorean identity: $\sin^2 x + \cos^2 x = 1$. So, you can use this identity to find $\cos x$ in terms of $\sin x$.

Substitute $\cos x = \sqrt{1 - \sin^2 x}$ into the equation:

$\sin x = \frac{-\sqrt{1 - \sin^2 x} \pm \sqrt{4(\cos^2 x + 612)}}{72}$