Question

Решить уравнение sin^2x-3sin-4=0

Answer 1

sin²x+3sinx-4=0

t=sinx

t²+3t-4=0

D=9+16=25 √D=5

t₁=(-3+5)/2=1

t₂=(-3-5)/2=4

sin x=1

x=(π/2)+2πn, n∈Z

Answer 2

To solve the equation $\sin^2x - 3\sin x - 4 = 0$, we can use some algebraic manipulation and trigonometric identities. Let's proceed with the solution:

Step 1: Factor the quadratic equation

Let $y = \sin x$. Then the equation becomes $y^2 - 3y - 4 = 0$. Now, we can factor the quadratic expression:

$(y - 4)(y + 1) = 0$

Step 2: Find the values of $y$

Setting each factor to zero and solving for $y$:

1. $y - 4 = 0$ $y = 4$

2. $y + 1 = 0$ $y = -1$

Step 3: Restore $x$ values using $\sin x = y$

For the first solution, $y = 4$, we have $\sin x = 4$. However, the sine function can only take values between -1 and 1, so there is no solution for this case.

For the second solution, $y = -1$, we have $\sin x = -1$.

Step 4: Find the values of $x$

Since $\sin x = -1$ at the angles $-\frac{\pi}{2}$ and $\frac{3\pi}{2}$, we get two solutions for $x$:

1. $x = -\frac{\pi}{2}$

2. $x = \frac{3\pi}{2}$

So, the solutions to the equation are $x = -\frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.