1. We are asked to solve the equation $$\sqrt{2} \sin 2x = \sin x - \cos x$$ exactly. 2. Recall the double-angle identity for sine: $$\sin 2x = 2 \sin x \cos x$$. 3. Substitute this into the equation: $$\sqrt{2} \cdot 2 \sin x \cos x = \sin x - \cos x$$ which simplifies to $$2 \sqrt{2} \sin x \cos x = \sin x - \cos x$$. 4. Rearrange all terms to one side: $$2 \sqrt{2} \sin x \cos x - \sin x + \cos x = 0$$. 5. Group terms: $$\sin x (2 \sqrt{2} \cos x - 1) + \cos x = 0$$. 6. Rewrite as: $$\sin x (2 \sqrt{2} \cos x - 1) = - \cos x$$. 7. Consider two cases: **Case 1:** $$\cos x = 0$$ - Then the equation becomes $$\sqrt{2} \sin 2x = \sin x - 0$$. - Since $$\cos x=0$$, $$x = \frac{\pi}{2} + k\pi$$ for integer $$k$$. - Check if these satisfy the original equation: - For $$x=\frac{\pi}{2}$$, $$\sin 2x = \sin \pi = 0$$, left side is 0. - Right side: $$\sin \frac{\pi}{2} - \cos \frac{\pi}{2} = 1 - 0 = 1$$. - Not equal, so no solution here. **Case 2:** $$\cos x \neq 0$$, divide both sides by $$\cos x$$: $$\sin x \left(2 \sqrt{2} - \frac{1}{\cos x}\right) = -1$$ This is complicated; instead, divide original equation by $$\cos x$$ (since $$\cos x \neq 0$$): $$\sqrt{2} \sin 2x = \sin x - \cos x$$ Divide both sides by $$\cos x$$: $$\sqrt{2} \frac{\sin 2x}{\cos x} = \frac{\sin x}{\cos x} - 1$$ Recall $$\sin 2x = 2 \sin x \cos x$$, so: $$\sqrt{2} \frac{2 \sin x \cos x}{\cos x} = \tan x - 1$$ Simplify: $$2 \sqrt{2} \sin x = \tan x - 1$$ But $$\tan x = \frac{\sin x}{\cos x}$$, so: $$2 \sqrt{2} \sin x = \frac{\sin x}{\cos x} - 1$$ Multiply both sides by $$\cos x$$: $$2 \sqrt{2} \sin x \cos x = \sin x - \cos x$$ This is the original equation again, so this approach loops. 8. Instead, rewrite the original equation as: $$\sqrt{2} \sin 2x - \sin x + \cos x = 0$$ Use $$\sin 2x = 2 \sin x \cos x$$: $$2 \sqrt{2} \sin x \cos x - \sin x + \cos x = 0$$ Group terms: $$\sin x (2 \sqrt{2} \cos x - 1) + \cos x = 0$$ Rewrite as: $$\sin x (2 \sqrt{2} \cos x - 1) = - \cos x$$ Divide both sides by $$\cos x$$ (assuming $$\cos x \neq 0$$): $$\tan x (2 \sqrt{2} \cos x - 1) = -1$$ Rewrite: $$2 \sqrt{2} \tan x \cos x - \tan x = -1$$ Since $$\tan x = \frac{\sin x}{\cos x}$$, $$\tan x \cos x = \sin x$$, so: $$2 \sqrt{2} \sin x - \tan x = -1$$ Rewrite $$\tan x$$ as $$\frac{\sin x}{\cos x}$$: $$2 \sqrt{2} \sin x - \frac{\sin x}{\cos x} = -1$$ Multiply both sides by $$\cos x$$: $$2 \sqrt{2} \sin x \cos x - \sin x = - \cos x$$ Rearranged: $$2 \sqrt{2} \sin x \cos x - \sin x + \cos x = 0$$ This is the original equation again, so this confirms the equation is symmetric. 9. Try substitution $$t = \tan x$$, use $$\sin x = \frac{t}{\sqrt{1+t^2}}$$ and $$\cos x = \frac{1}{\sqrt{1+t^2}}$$. Then: $$\sqrt{2} \sin 2x = \sqrt{2} \cdot 2 \sin x \cos x = 2 \sqrt{2} \cdot \frac{t}{\sqrt{1+t^2}} \cdot \frac{1}{\sqrt{1+t^2}} = \frac{2 \sqrt{2} t}{1+t^2}$$ Right side: $$\sin x - \cos x = \frac{t}{\sqrt{1+t^2}} - \frac{1}{\sqrt{1+t^2}} = \frac{t - 1}{\sqrt{1+t^2}}$$ Equation becomes: $$\frac{2 \sqrt{2} t}{1+t^2} = \frac{t - 1}{\sqrt{1+t^2}}$$ Multiply both sides by $$ (1+t^2) \sqrt{1+t^2} $$: $$2 \sqrt{2} t \sqrt{1+t^2} = (t - 1)(1+t^2)$$ Square both sides to eliminate the square root: $$4 \cdot 2 \cdot t^2 (1+t^2) = (t - 1)^2 (1+t^2)^2$$ Simplify left side: $$8 t^2 (1+t^2) = (t - 1)^2 (1+t^2)^2$$ Divide both sides by $$ (1+t^2) $$ (assuming $$t \neq \pm i$$): $$8 t^2 = (t - 1)^2 (1+t^2)$$ Expand right side: $$(t - 1)^2 = t^2 - 2t + 1$$ So: $$8 t^2 = (t^2 - 2t + 1)(1 + t^2) = (t^2 - 2t + 1)(1 + t^2)$$ Expand: $$= t^2(1 + t^2) - 2t(1 + t^2) + 1(1 + t^2) = t^2 + t^4 - 2t - 2t^3 + 1 + t^2$$ Combine like terms: $$t^4 - 2 t^3 + 2 t^2 - 2 t + 1$$ So the equation is: $$8 t^2 = t^4 - 2 t^3 + 2 t^2 - 2 t + 1$$ Bring all terms to one side: $$0 = t^4 - 2 t^3 + 2 t^2 - 2 t + 1 - 8 t^2 = t^4 - 2 t^3 - 6 t^2 - 2 t + 1$$ 10. Solve quartic: $$t^4 - 2 t^3 - 6 t^2 - 2 t + 1 = 0$$ Try rational roots: possible roots $$\pm 1$$. Test $$t=1$$: $$1 - 2 - 6 - 2 + 1 = -8 \neq 0$$ Test $$t=-1$$: $$1 + 2 - 6 + 2 + 1 = 0$$ So $$t = -1$$ is a root. Divide polynomial by $$t + 1$$: Using polynomial division or synthetic division: Coefficients: 1, -2, -6, -2, 1 Divide by $$t + 1$$: Bring down 1 Multiply by -1: -1 Add to -2: -3 Multiply by -1: 3 Add to -6: -3 Multiply by -1: 3 Add to -2: 1 Multiply by -1: -1 Add to 1: 0 Quotient: $$t^3 - 3 t^2 - 3 t + 1$$ 11. Solve cubic: $$t^3 - 3 t^2 - 3 t + 1 = 0$$ Try rational roots $$\pm 1$$: For $$t=1$$: $$1 - 3 - 3 + 1 = -4 \neq 0$$ For $$t=-1$$: $$-1 - 3 + 3 + 1 = 0$$ So $$t = -1$$ is again a root. Divide cubic by $$t + 1$$: Coefficients: 1, -3, -3, 1 Bring down 1 Multiply by -1: -1 Add to -3: -4 Multiply by -1: 4 Add to -3: 1 Multiply by -1: -1 Add to 1: 0 Quotient: $$t^2 - 4 t + 1$$ 12. Solve quadratic: $$t^2 - 4 t + 1 = 0$$ Use quadratic formula: $$t = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2 \sqrt{3}}{2} = 2 \pm \sqrt{3}$$ 13. So the roots are: $$t = -1$$ (double root), $$t = 2 + \sqrt{3}$$, and $$t = 2 - \sqrt{3}$$. 14. Recall $$t = \tan x$$, so: $$\tan x = -1$$ or $$\tan x = 2 + \sqrt{3}$$ or $$\tan x = 2 - \sqrt{3}$$. 15. Write general solutions: - For $$\tan x = -1$$: $$x = \arctan(-1) + k \pi = -\frac{\pi}{4} + k \pi$$ - For $$\tan x = 2 + \sqrt{3}$$: $$x = \arctan(2 + \sqrt{3}) + k \pi$$ - For $$\tan x = 2 - \sqrt{3}$$: $$x = \arctan(2 - \sqrt{3}) + k \pi$$ where $$k$$ is any integer. **Final answer:** $$x = -\frac{\pi}{4} + k \pi, \quad x = \arctan(2 + \sqrt{3}) + k \pi, \quad x = \arctan(2 - \sqrt{3}) + k \pi, \quad k \in \mathbb{Z}$$