Question

Cos(pi/2 + 2x)=sqrt(2)*sin(x)

Answer 1

-sin2x=√2*sinx
-2sinxcosx-√2*sinx=0
-sinx(2cosx+√2)=0
1)-sinx=0
sinx=0
x=Pi*n , nĕZ
2)2cosx=-√2
cosx=-√2/2
x=±Pi/4+2Pin.

Answer 2

To solve the trigonometric equation cos(pi/2 + 2x) = sqrt(2) * sin(x), we can use trigonometric identities and algebraic manipulations. Let's start by simplifying the left-hand side of the equation:

cos(pi/2 + 2x) = cos(pi/2) * cos(2x) - sin(pi/2) * sin(2x) = 0 * cos(2x) - 1 * sin(2x) = -sin(2x)

Now, the equation becomes:

-sin(2x) = sqrt(2) * sin(x)

Next, we'll use the double-angle formula for sine:

sin(2x) = 2 * sin(x) * cos(x)

Substitute this back into the equation:

-2 * sin(x) * cos(x) = sqrt(2) * sin(x)

Now, to solve for sin(x), we can divide both sides by sin(x) (assuming sin(x) ≠ 0):

-2 * cos(x) = sqrt(2)

Now, divide both sides by -2:

cos(x) = -sqrt(2) / 2

We know that cos(π/4) = sqrt(2) / 2, and cos(π/4) is negative in the second and third quadrants. So, we can write:

cos(x) = cos(π/4)

Now, for the general solution for x, we can add multiples of the period (2π) to the angle in the interval [0, 2π]:

x = π/4 + 2πn or x = 3π/4 + 2πn

where 'n' is an integer representing all possible solutions.

Thus, the solutions to the equation are:

x = π/4 + 2πn or x = 3π/4 + 2πn

where 'n' is an integer.