Question

1/2+cos^2•3x/2-sin^2•3x/2=0

Answer 1

$\frac{1}{2} + { \cos( \frac{3x}{2} ) }^{2} - { \sin( \frac{3x}{2} ) }^{2} = 0 \\ \ { \sin(x) }^{2} = 1 - { \cos(x) }^{2} \\ \frac{1}{2} + { \cos( \frac{3x}{2} ) }^{2} - 1 + { \cos( \frac{3x}{2} ) }^{2} = 0 \\ 2{ \cos( \frac{3x}{2} ) }^{2} = \frac{1}{2} \\ { \cos( \frac{3x}{2} ) }^{2} = \frac{1}{4} \\ \cos( \frac{3x}{2} ) = \frac{1}{2} \\ \frac{3x}{2} = + - \frac{\pi}{3} + 2\pi \times n \\ 3x = + - \frac{2\pi}{3} + 4\pi \times n \\ x = + - \frac{2\pi}{9} + \frac{4\pi \times n}{3}$

Answer 2

To simplify the given expression, let's first write it with clearer notation:

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

Now, we can work on simplifying the expression step by step:

Step 1: Combine the fractions on the left-hand side: (1/2) + (cos^2(3x) - sin^2(3x))/2 = 0

Step 2: Use the trigonometric identity cos^2θ - sin^2θ = cos(2θ):

(1/2) + cos(2(3x))/2 = 0

Step 3: Multiply the entire equation by 2 to eliminate the fraction:

1 + cos(2(3x)) = 0

Step 4: Now, isolate the cos(2(3x)) term:

cos(2(3x)) = -1

Step 5: Recall that the cosine function takes on the value of -1 at odd multiples of π (pi):

2(3x) = π + 2nπ, where n is an integer.

Step 6: Solve for x:

3x = π/2 + nπ

x = (π/6) + (nπ/3)

So, the general solution for the given equation is:

x = (π/6) + (nπ/3), where n is an integer.