Question

Sin^2(3x)=3cos^2(3x)

Answer 1

$\frac{1 - \cos(6x) }{2} = \frac{3(1 + \cos(6x)) }{2}$
$3 + 3 \cos(6x) - 1 + \cos(6x) = 0$
$4 \cos(6x) = - 2$
$\cos(6x) = - \frac{1}{2}$
$6x = + - \frac{2}{3} \pi + 2\pi \times k$
$x = + - \frac{\pi}{9} + \frac{\pi \times k}{3}$

Answer 2

To solve the trigonometric equation sin^2(3x) = 3cos^2(3x), we can use the Pythagorean identity: sin^2(theta) + cos^2(theta) = 1. Rearranging the equation, we get:

sin^2(3x) - 3cos^2(3x) = 1 - 3cos^2(3x)

Now, we can substitute the identity sin^2(theta) = 1 - cos^2(theta):

(1 - cos^2(3x)) - 3cos^2(3x) = 1 - 3cos^2(3x)

Now, simplify:

1 - cos^2(3x) - 3cos^2(3x) = 1 - 3cos^2(3x)

Combine like terms:

1 - 4cos^2(3x) = 1 - 3cos^2(3x)

Now, bring all the terms to one side of the equation:

1 - 1 - 4cos^2(3x) + 3cos^2(3x) = 0

Simplify further:

-4cos^2(3x) + 3cos^2(3x) = 0

Now, combine like terms again:

-cos^2(3x) = 0

Finally, divide both sides by -1:

cos^2(3x) = 0

Now, we have a simpler equation to solve. To find the solutions for x, we need to consider when the cosine of 3x is equal to zero.

cos^2(3x) = 0

Taking the square root of both sides:

cos(3x) = 0

Now, we are looking for values of 3x where the cosine is equal to zero. The cosine function is zero at odd multiples of pi/2, so:

3x = (2n + 1) * pi/2

where n is an integer.

Now, solve for x:

x = (2n + 1) * pi/6

So the solutions for x are:

x = pi/6, 5pi/6, 3pi/2, 7pi/6, 11pi/6, ...

These are the values of x that satisfy the equation sin^2(3x) = 3cos^2(3x).