Question

Решите sin^2x+3cosx=3

Answer 1

Смотри.........................

Answer 2

sin^2x+3cosx-3=0              

sin^2+cos^2=1    sin^2x=1-cos^2x

1-cos^2x+3cosx-3=0

 -cos^2x+3cosx-2=0

 пусть cosx=t

-t^2+3t-2=0 /*(-1)

t^2-3t+2=0

D = b2 - 4ac

D = 9 - 8 = 1

t1,2 = -b ± √D/2a

t1 = 3 + 1/2 = 4/2 = 2

t2 = 3 - 1/2 = 2/2 = 1

cosx=2 не подходить так как окружност от -1 до 1

cosx=1

x=2пиn

Ответ: x=2пиn             n типо целое

Answer 3

To solve the equation sin^2(x) + 3cos(x) = 3, we'll need to use trigonometric identities to manipulate the equation and find the solutions for x.

Let's start by using the Pythagorean identity: sin^2(x) + cos^2(x) = 1.

Rearranging the given equation, we get: sin^2(x) + 3cos(x) - 3 = 0

Now, let's use the Pythagorean identity to replace sin^2(x) in terms of cos(x): 1 - cos^2(x) + 3cos(x) - 3 = 0

Now, this equation is quadratic in terms of cos(x). Let's solve it by setting it equal to zero:

cos^2(x) - 3cos(x) + 2 = 0

Now, we can factor this quadratic equation:

(cos(x) - 2)(cos(x) - 1) = 0

Now, we set each factor to zero and solve for cos(x):

1. cos(x) - 2 = 0 cos(x) = 2 [Discard this solution, as the cosine function's range is between -1 and 1]

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

Now, we have found the possible values of cos(x). To find the corresponding values of x, we take the inverse cosine (arccos) of each solution:

1. cos(x) = 1 x = arccos(1) = 0

So, the solution to the equation sin^2(x) + 3cos(x) = 3 is x = 0.