Solve the equation 2x - 52 x = (1)/(9)

We know that $\cos 2x = 1 - 2\sin^2 x$. Substitute this into the equation:

$$(1 - 2\sin^2 x) - 5\sin^2 x = \frac{1}{9}$$

Combine the $\sin^2 x$ terms:

$$1 - 7\sin^2 x = \frac{1}{9}$$

Isolate the $\sin^2 x$ term:

$$7\sin^2 x = 1 - \frac{1}{9}$$ $$7\sin^2 x = \frac{9}{9} - \frac{1}{9}$$ $$7\sin^2 x = \frac{8}{9}$$

Divide by 7:

$$\sin^2 x = \frac{8}{63}$$

Take the square root of both sides:

$$\sin x = \pm\sqrt{\frac{8}{63}}$$

Calculate the decimal value:

$$\sin x \approx \pm\sqrt{0.126984}$$ $$\sin x \approx \pm 0.356348$$

Let $\alpha$ be the acute angle such that $\sin \alpha = 0.356348$.

$$\alpha = \arcsin(0.356348)$$ $$\alpha \approx 20.87^\circ$$

Case 1: $\sin x = 0.356348$ (positive)
Since $\sin x$ is positive, $x$ lies in Quadrant I or Quadrant II.

• In Quadrant I: $x_1 = \alpha \approx 20.87^\circ$
• In Quadrant II: $x_2 = 180^\circ - \alpha \approx 180^\circ - 20.87^\circ = 159.13^\circ$

Case 2: $\sin x = -0.356348$ (negative)
Since $\sin x$ is negative, $x$ lies in Quadrant III or Quadrant IV.

• In Quadrant III: $x_3 = 180^\circ + \alpha \approx 180^\circ + 20.87^\circ = 200.87^\circ$
• In Quadrant IV: $x_4 = 360^\circ - \alpha \approx 360^\circ - 20.87^\circ = 339.13^\circ$

• $x_1 \approx 21^\circ$
• $x_2 \approx 159^\circ$
• $x_3 \approx 201^\circ$
• $x_4 \approx 339^\circ$

The solutions are $21^\circ, 159^\circ, 201^\circ, 339^\circ$.