Question

Cos(2x-(3п)/2)=√(2)sinx

Answer 1

Cos(2x-(3п)/2)=√(2)sinx
Cos((3п)/2-2x)=√(2)sinx
-sin2x-√2sinx=0
-2sinxcosx-√2sinx=0
-sinx(2cosx+√2)=0
sinx=0⇒x=πk,k∈z
2cosx+√2=0⇒cosx=-√2/2⇒x=+-3π/4+2πk,k∈z

Answer 2

To solve the trigonometric equation cos(2x - 3π/2) = √2sinx, we'll first manipulate the equation to get it in terms of a single trigonometric function. Then, we'll solve for x.

Step 1: Use the trigonometric identity cos(2x) = 1 - 2sin^2(x) to simplify the left side of the equation.

cos(2x - 3π/2) = cos(2x)cos(3π/2) + sin(2x)sin(3π/2) cos(2x - 3π/2) = -sin(2x)

Step 2: Replace cos(2x - 3π/2) with -sin(2x) in the equation:

-sin(2x) = √2sinx

Step 3: Move all the terms to one side of the equation:

-sin(2x) - √2sinx = 0

Step 4: Factor out sinx from the expression:

sinx(-2cosx - √2) = 0

Step 5: Now we have two cases to consider:

Case 1: sinx = 0 If sinx = 0, then x can take values of πn, where n is an integer.

Case 2: -2cosx - √2 = 0 Solving for cosx:

-2cosx = √2 cosx = -√2 / 2

The angle whose cosine is -√2/2 is 3π/4 radians. However, we need to consider both the positive and negative values of cosx, which means we have two solutions for this case:

x = 3π/4 or x = -3π/4

So the complete solution for the equation cos(2x - 3π/2) = √2sinx is:

x = πn, 3π/4 + 2πn, or -3π/4 + 2πn, where n is an integer.