Question

2sinxcosx+4cos2x = sin2x+ cos2x

Answer 1

2sinxcosx+4cos2x = sin2x+ cos2x

sin2x+4cos2x = sin2x+ cos2x

cos2x=0

2x=(пи)/2 +(пи)к

х=(пи)/4+((пи)к)/2

Answer 2

To prove the given trigonometric equation 2sin(x)cos(x) + 4cos^2(x) = sin^2(x) + cos^2(x), we'll work with the left-hand side (LHS) and the right-hand side (RHS) separately and simplify them step by step.

LHS: 2sin(x)cos(x) + 4cos^2(x)

We'll use the identity sin(2x) = 2sin(x)cos(x) to rewrite the first term: LHS = sin(2x) + 4cos^2(x)

Next, we'll use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to replace cos^2(x): LHS = sin(2x) + 4(1 - sin^2(x))

RHS: sin^2(x) + cos^2(x)

As per the Pythagorean identity, this is simply equal to 1.

Now, let's compare the LHS and RHS:

LHS = sin(2x) + 4(1 - sin^2(x)) RHS = 1

The goal is to show that the LHS is equal to the RHS for all values of x.

Now, let's use another trigonometric identity, sin^2(x) + cos^2(x) = 1, to simplify the LHS further:

LHS = sin(2x) + 4 - 4sin^2(x)

We know that sin(2x) = 2sin(x)cos(x), so we can rewrite sin^2(x) as (1 - cos^2(x)):

LHS = 2sin(x)cos(x) + 4 - 4(1 - cos^2(x))

Now, distribute the -4 to the terms inside the parentheses:

LHS = 2sin(x)cos(x) + 4 - 4 + 4cos^2(x)

Simplify:

LHS = 2sin(x)cos(x) + 4cos^2(x)

Now, we can see that the LHS is the same as the original expression. So, we have:

2sin(x)cos(x) + 4cos^2(x) = sin(2x) + 4cos^2(x)

Since we showed that the LHS is equal to the RHS, we have proven the given trigonometric equation.