1. **State the problem:** Solve the equation $\left(c \sin x + \cos x\right)^2 = 2$ for $x$. 2. **Recall the formula and rules:** The square of a sum is expanded as $\left(a+b\right)^2 = a^2 + 2ab + b^2$. 3. **Expand the left side:** $$\left(c \sin x + \cos x\right)^2 = c^2 \sin^2 x + 2c \sin x \cos x + \cos^2 x$$ 4. **Rewrite the equation:** $$c^2 \sin^2 x + 2c \sin x \cos x + \cos^2 x = 2$$ 5. **Use the Pythagorean identity:** $$\sin^2 x + \cos^2 x = 1$$ 6. **Substitute $\cos^2 x = 1 - \sin^2 x$:** $$c^2 \sin^2 x + 2c \sin x \cos x + 1 - \sin^2 x = 2$$ 7. **Group like terms:** $$\left(c^2 - 1\right) \sin^2 x + 2c \sin x \cos x + 1 = 2$$ 8. **Subtract 2 from both sides:** $$\left(c^2 - 1\right) \sin^2 x + 2c \sin x \cos x + 1 - 2 = 0$$ 9. **Simplify:** $$\left(c^2 - 1\right) \sin^2 x + 2c \sin x \cos x - 1 = 0$$ 10. **Rewrite as:** $$\left(c^2 - 1\right) \sin^2 x + 2c \sin x \cos x = 1$$ 11. **Divide both sides by $\cos^2 x$ (assuming $\cos x \neq 0$):** $$\cancel{\cos^2 x} \left(\frac{c^2 - 1}{\cancel{\cos^2 x}} \sin^2 x + 2c \frac{\sin x}{\cos x} \right) = \frac{1}{\cos^2 x}$$ This simplifies to: $$\left(c^2 - 1\right) \tan^2 x + 2c \tan x = \sec^2 x$$ 12. **Use the identity $\sec^2 x = 1 + \tan^2 x$:** $$\left(c^2 - 1\right) \tan^2 x + 2c \tan x = 1 + \tan^2 x$$ 13. **Bring all terms to one side:** $$\left(c^2 - 1\right) \tan^2 x - \tan^2 x + 2c \tan x - 1 = 0$$ 14. **Simplify:** $$\left(c^2 - 2\right) \tan^2 x + 2c \tan x - 1 = 0$$ 15. **Let $t = \tan x$, then solve the quadratic:** $$\left(c^2 - 2\right) t^2 + 2c t - 1 = 0$$ 16. **Use quadratic formula:** $$t = \frac{-2c \pm \sqrt{(2c)^2 - 4 (c^2 - 2)(-1)}}{2 (c^2 - 2)} = \frac{-2c \pm \sqrt{4c^2 + 4(c^2 - 2)}}{2 (c^2 - 2)}$$ 17. **Simplify under the square root:** $$4c^2 + 4c^2 - 8 = 8c^2 - 8 = 8(c^2 - 1)$$ 18. **Final expression for $t$:** $$t = \frac{-2c \pm 2\sqrt{2(c^2 - 1)}}{2 (c^2 - 2)} = \frac{-c \pm \sqrt{2(c^2 - 1)}}{c^2 - 2}$$ 19. **Therefore, the solutions for $x$ are:** $$x = \arctan\left(\frac{-c + \sqrt{2(c^2 - 1)}}{c^2 - 2}\right) + k\pi \quad \text{or} \quad x = \arctan\left(\frac{-c - \sqrt{2(c^2 - 1)}}{c^2 - 2}\right) + k\pi, \quad k \in \mathbb{Z}$$ 20. **Note:** Solutions exist only if the discriminant $2(c^2 - 1) \geq 0$, i.e., $c^2 \geq 1$. **Final answer:** $$x = \arctan\left(\frac{-c \pm \sqrt{2(c^2 - 1)}}{c^2 - 2}\right) + k\pi, \quad k \in \mathbb{Z}$$