Question

Решить: cos^2x+3sinx-3=0 , [-2p; 3p]

Answer 1

task/29816879  решить  cos²x +3sinx - 3 =0 ;   x ∈ [-2π ; 3π]

решение. cos²x +3sinx-3 =0⇔1 -sin²x +3sinx-3 =0 ⇔sin²x-3sinx+2 =0                                                             [ sinx =2 ; sinx =1 .    но  sinx =2 >  1→ посторонний корень ;

sinx = 1  ⇒ x =  π/2 +2πn , n ∈ ℤ    при n = - 1 ; 0 ; 1  получаем

ответ.   - 3π/2 ; π/2 ;  5π/2 .  

* * * - 2π ≤ π/2 +2πn ≤  3π ⇔ -2 ≤ 1/2+2n ≤ 3 ⇔ -2,5 ≤ 2n ≤ 2,5 ⇔

- 1,25 ≤ n ≤ 1,25  ⇒ n = - 1 ; 0 ; 1  * * *                                        

Answer 2

To solve the equation cos^2(x) + 3sin(x) - 3 = 0 over the interval [-2π, 3π], follow these steps:

Step 1: Make the substitution cos^2(x) = 1 - sin^2(x) into the equation to obtain a quadratic equation in sin(x):

1 - sin^2(x) + 3sin(x) - 3 = 0.

Step 2: Rearrange the equation:

-sin^2(x) + 3sin(x) - 2 = 0.

Step 3: Multiply the equation by -1 to get a positive coefficient for the sin^2(x) term:

sin^2(x) - 3sin(x) + 2 = 0.

Step 4: Factor the quadratic equation:

(sin(x) - 2)(sin(x) - 1) = 0.

Step 5: Now, set each factor equal to zero and solve for sin(x):

1. sin(x) - 2 = 0 sin(x) = 2 (This solution is not possible since the sine function's range is [-1, 1].)

2. sin(x) - 1 = 0 sin(x) = 1

Step 6: The solution sin(x) = 1 is possible and occurs when x is π/2 (90 degrees).

Step 7: Check if there are any additional solutions within the specified interval [-2π, 3π]. For this, you can calculate the values of sin(x) and cos(x) at the endpoints of the interval:

At x = -2π: sin(-2π) = 0 and cos(-2π) = 1. At x = 3π: sin(3π) = 0 and cos(3π) = -1.

Since there are no solutions between -2π and 3π other than x = π/2, the final solution for the equation cos^2(x) + 3sin(x) - 3 = 0 over the interval [-2π, 3π] is:

x = π/2.