Question

Solve the equation. (Find all solutions of the equation in the interval [0, 2pi). Enter your answers as a comma-separated list.)
(3 sin(2x) + 3 cos(2x))^2=9

Answer 1

I got

$0.75 + 2\pi \times n$

$5.28 + 2\pi \times n$

$1 + 2\pi \times n$

5.53+2pi times n

WORK:

$(3 \sin(2x) + 3 \cos(2x) )) {}^{2} = 9$

$\sin(2x) + \cos(2x) {}^{2} = 3$

$\sin(x) + \cos(4 {x}^{2} ) = 3$

$4 {x}^{2} + x = 3$

$4 {x}^{2} + x - 3 = 0$

$(4x {}^{2} + 4x - 3x - 3 = 0$

$4x(x + 1) - 3(x + 1) = 0$

$(4x - 3)(x + 1)$

Solve for x

$4x - 3 = 0$

$4x = 3$

$x = \frac{3}{4}$

$x + 1 = 0$

$x = - 1$

While -1 is a solution, it isn't within the interval of 0,2pi.

However, we can use the reference angle identity

$\cos(x) = \cos(x + 2\pi)$

So some more possible points are

$- 1 + 2\pi$

Since it on a interval, it must be less than 2 pi or 6.28.

Let find some more points

$- 1 + 2\pi = 5.28$

So 5.28 is a point,

$\frac{3}{4} + 2\pi = 7.03$

That is too big since -6.28 is smaller than 7.03

We can use the identity

$\cos(x) = \cos( - x)$

$\cos( - 1) = \cos( 1 )$

1 is smaller than 2pi so it is true.

$\cos( \frac{3}{4} ) = \cos( - \frac{3}{4} )$

Negative 3/4 is smaller than zero so it not a solution.

$0.75$

$5.28$

$1$

Make sure to include

$2\pi \times n$

so

$0.75 + 2\pi \times n$

$5.28 + 2\pi \times n$

$1 + 2\pi \times n$

We can also -3/4 plus 2 pi.

Which is about 6

5.53.