Question

Sin^2x-Sinx-2=0 tenglamani eching

Answer 1

$sin^2x-sinx-2=0 \\ (sinx-2)(sinx+1)=0$
решения не имеет x∈∅

Область допустимых значений sinx=[-1;1]

Answer 2

I can help you solve this equation. First, let's rewrite it as a quadratic equation in terms of sin x:

sin^2 x + sin x - 2 = 0

Now, we can use the quadratic formula to find the values of sin x that satisfy the equation:

sin x = (-1 ± √(1 + 8))/2

sin x = (-1 + √9)/2 or sin x = (-1 - √9)/2

sin x = 1 or sin x = -2

The second value of sin x is not possible, since the sine function can only range from -1 to 1. Therefore, we only consider the first value of sin x:

sin x = 1

To find the values of x that make this true, we can use the inverse sine function:

x = arcsin(1)

x = π/2 + 2kπ, where k is any integer

This is the general solution of the equation. If we want to find the specific solutions in a given interval, we can substitute different values of k and see which ones fall within the interval. For example, if we want to find the solutions in the interval [0, 2π], we can try k = 0, 1, -1, etc. and see that only k = 0 gives a valid solution:

x = π/2 + 2(0)π

x = π/2

This is the only solution in the interval [0, 2π]. If we want to find the solutions in a different interval, we can repeat the same process with different values of k.

I hope this helps you understand how to solve this equation. If you have any questions, feel free to ask me.