1. **State the problem:** Solve the trigonometric equation $$6 \tan x \sin x = 5 \sin x - \cos x$$ for $$0^\circ \leq x \leq 360^\circ$$. 2. **Rewrite the equation:** Recall that $$\tan x = \frac{\sin x}{\cos x}$$, so substitute: $$6 \cdot \frac{\sin x}{\cos x} \cdot \sin x = 5 \sin x - \cos x$$ 3. **Simplify the left side:** $$6 \frac{\sin^2 x}{\cos x} = 5 \sin x - \cos x$$ 4. **Multiply both sides by $$\cos x$$ to clear the denominator:** $$6 \sin^2 x = 5 \sin x \cos x - \cos^2 x$$ 5. **Bring all terms to one side:** $$6 \sin^2 x - 5 \sin x \cos x + \cos^2 x = 0$$ 6. **Use the Pythagorean identity $$\sin^2 x + \cos^2 x = 1$$ to express $$\sin^2 x$$:** $$6 (1 - \cos^2 x) - 5 \sin x \cos x + \cos^2 x = 0$$ 7. **Expand and simplify:** $$6 - 6 \cos^2 x - 5 \sin x \cos x + \cos^2 x = 0$$ $$6 - 5 \cos^2 x - 5 \sin x \cos x = 0$$ 8. **Rewrite:** $$6 = 5 \cos^2 x + 5 \sin x \cos x$$ 9. **Divide both sides by 5:** $$\frac{6}{5} = \cos^2 x + \sin x \cos x$$ 10. **Factor the right side:** $$\cos x (\cos x + \sin x) = \frac{6}{5}$$ 11. **Note that $$\cos x$$ and $$\cos x + \sin x$$ are bounded between -1 and 1, so check possible values:** 12. **Try to solve by substitution or numerical methods:** 13. **Alternatively, return to step 4 and consider cases:** - Case 1: $$\sin x = 0$$ Substitute into original equation: $$6 \tan x \sin x = 0$$ and $$5 \sin x - \cos x = -\cos x$$ So equation becomes $$0 = -\cos x$$, which implies $$\cos x = 0$$. But $$\sin x = 0$$ and $$\cos x = 0$$ cannot be true simultaneously. So no solution here. - Case 2: $$\sin x \neq 0$$, divide original equation by $$\sin x$$: $$6 \tan x = 5 - \frac{\cos x}{\sin x} = 5 - \cot x$$ 14. **Recall $$\tan x = \frac{\sin x}{\cos x}$$ and $$\cot x = \frac{\cos x}{\sin x}$$, so rewrite:** $$6 \tan x + \cot x = 5$$ 15. **Multiply both sides by $$\sin x \cos x$$ to clear denominators:** $$6 \sin x \cos x + \cos^2 x = 5 \sin x \cos x$$ 16. **Bring all terms to one side:** $$6 \sin x \cos x + \cos^2 x - 5 \sin x \cos x = 0$$ $$\cos^2 x + (6 - 5) \sin x \cos x = 0$$ $$\cos^2 x + \sin x \cos x = 0$$ 17. **Factor $$\cos x$$:** $$\cos x (\cos x + \sin x) = 0$$ 18. **Set each factor to zero:** - $$\cos x = 0$$ - $$\cos x + \sin x = 0$$ 19. **Solve $$\cos x = 0$$ for $$0^\circ \leq x \leq 360^\circ$$:** $$x = 90^\circ, 270^\circ$$ 20. **Solve $$\cos x + \sin x = 0$$:** $$\sin x = -\cos x$$ Divide both sides by $$\cos x$$ (where $$\cos x \neq 0$$): $$\tan x = -1$$ 21. **Find $$x$$ such that $$\tan x = -1$$ in $$0^\circ \leq x \leq 360^\circ$$:** $$x = 135^\circ, 315^\circ$$ 22. **Check all solutions in original equation:** - For $$x=90^\circ$$, $$\tan 90^\circ$$ is undefined, discard. - For $$x=270^\circ$$, $$\tan 270^\circ$$ is undefined, discard. - For $$x=135^\circ$$ and $$315^\circ$$, $$\tan x$$ is defined. 23. **Verify $$x=135^\circ$$:** $$6 \tan 135^\circ \sin 135^\circ = 6 (-1)(\frac{\sqrt{2}}{2}) = -3 \sqrt{2}$$ $$5 \sin 135^\circ - \cos 135^\circ = 5 (\frac{\sqrt{2}}{2}) - (-\frac{\sqrt{2}}{2}) = 3 \sqrt{2}$$ Not equal, so discard. 24. **Verify $$x=315^\circ$$:** $$6 \tan 315^\circ \sin 315^\circ = 6 ( -1)(-\frac{\sqrt{2}}{2}) = 3 \sqrt{2}$$ $$5 \sin 315^\circ - \cos 315^\circ = 5 (-\frac{\sqrt{2}}{2}) - (\frac{\sqrt{2}}{2}) = -3 \sqrt{2}$$ Not equal, discard. 25. **Re-examine step 13 case 1:** If $$\sin x = 0$$, then original equation becomes: $$6 \tan x \cdot 0 = 5 \cdot 0 - \cos x$$ $$0 = -\cos x$$ $$\cos x = 0$$ No $$x$$ satisfies both $$\sin x=0$$ and $$\cos x=0$$ simultaneously. 26. **Re-examine step 14:** Equation $$6 \tan x + \cot x = 5$$ can be rewritten as: $$6 \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = 5$$ Multiply both sides by $$\sin x \cos x$$: $$6 \sin^2 x + \cos^2 x = 5 \sin x \cos x$$ 27. **Use $$\sin^2 x = 1 - \cos^2 x$$:** $$6 (1 - \cos^2 x) + \cos^2 x = 5 \sin x \cos x$$ $$6 - 6 \cos^2 x + \cos^2 x = 5 \sin x \cos x$$ $$6 - 5 \cos^2 x = 5 \sin x \cos x$$ 28. **Bring all terms to one side:** $$6 - 5 \cos^2 x - 5 \sin x \cos x = 0$$ 29. **Divide by 5:** $$\frac{6}{5} - \cos^2 x - \sin x \cos x = 0$$ 30. **Rewrite:** $$\cos^2 x + \sin x \cos x = \frac{6}{5}$$ 31. **Note that $$\cos^2 x + \sin x \cos x \leq 1 + 1 = 2$$ but $$\frac{6}{5} = 1.2$$, so solutions may exist. 32. **Try substitution $$t = \cos x$$ and $$s = \sin x$$ with $$s^2 + t^2 = 1$$:** Equation becomes: $$t^2 + s t = 1.2$$ 33. **Express $$s = \pm \sqrt{1 - t^2}$$ and substitute:** $$t^2 + t (\pm \sqrt{1 - t^2}) = 1.2$$ 34. **Solve for $$t$$ numerically or graphically.** **Final answer:** The exact solutions are the $$x$$ values satisfying $$\cos x (\cos x + \sin x) = \frac{6}{5}$$ within $$0^\circ \leq x \leq 360^\circ$$. Since $$\frac{6}{5} > 1$$, and $$\cos x (\cos x + \sin x) \leq 1$$, no real solutions exist. **Therefore, the only solutions come from:** $$\cos x (\cos x + \sin x) = 0$$ which gives $$x = 90^\circ, 270^\circ$$ (discarded due to undefined $$\tan x$$) and $$x = 135^\circ, 315^\circ$$ (which do not satisfy original equation). Hence, **no solutions** in $$0^\circ \leq x \leq 360^\circ$$. --- **Summary:** - No valid $$x$$ satisfies the equation in the given interval.