Question

2sin^2(3pi/2-x)-3sin(pi/2+x)-2=0

Answer 1

$2\sin ^2( \frac{3 \pi }{2} -x)-3\sin ( \frac{\pi}{2} +x)-2=0 \\ 2\cos^2x-3\cos x-2=0$
пусть cos x=t (|t|≤1) получаем
2t²-3t-2=0
D=b²-4ac=9+16=25
t1=(3+5)/4=2 - ∉ (|t|≤1)
t2=(3-5)/4=-1/2
Возвращаемся к замене
$\cos x=- \frac{1}{2} \\ x=\pm \frac{2 \pi }{3} +2 \pi n,n \in Z$

Answer 2

To solve the equation $2\sin^2\left(\frac{3\pi}{2}-x\right)-3\sin\left(\frac{\pi}{2}+x\right)-2=0$, we can use trigonometric identities to simplify it.

Let's start by making some substitutions to simplify the equation:

1. Use the identity $\sin(\pi - \theta) = \sin(\theta)$ to rewrite $\sin\left(\frac{3\pi}{2} - x\right)$ as $\sin\left(\frac{\pi}{2} + x\right)$: $2\sin^2\left(\frac{\pi}{2} + x\right)-3\sin\left(\frac{\pi}{2} + x\right)-2=0$

2. Define a new variable $y = \sin\left(\frac{\pi}{2} + x\right)$ so that the equation becomes: $2y^2 - 3y - 2 = 0$

Now, you have a quadratic equation in $y$. To solve for $y$, you can use the quadratic formula:

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

In this case, $a = 2$, $b = -3$, and $c = -2$. Plugging these values into the formula:

$y = \frac{3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)}$

$y = \frac{3 \pm \sqrt{17}}{4}$

So, you have two possible values for $y$:

1. $y = \frac{3 + \sqrt{17}}{4}$
2. $y = \frac{3 - \sqrt{17}}{4}$

Now that you have the values for $y$, you can find $x$:

For $y = \frac{3 + \sqrt{17}}{4}$: $\sin\left(\frac{\pi}{2} + x\right) = \frac{3 + \sqrt{17}}{4}$

Using inverse sine (arcsin) to find $x$:

\frac{\pi}{2} + x = \arcsin\left(\frac{3 + \sqrt{17}}{4}

x = \arcsin\left(\frac{3 + \sqrt{17}}{4} - $\frac{\pi}{2}$

For $y = \frac{3 - \sqrt{17}}{4}$: $\sin\left(\frac{\pi}{2} + x\right) = \frac{3 - \sqrt{17}}{4}$

Using inverse sine (arcsin) to find $x$:

\frac{\pi}{2} + x = \arcsin\left(\frac{3 - \sqrt{17}}{4}

x = \arcsin\left(\frac{3 - \sqrt{17}}{4} - $\frac{\pi}{2}$

These are the solutions for $x$ that satisfy the equation $2\sin^2\left(\frac{3\pi}{2}-x\right)-3\sin\left(\frac{\pi}{2}+x\right)-2=0$