Solve Trig Equation C83E70

1. **State the problem:** Solve the equation $$3\sin^2 2peter + 8\cos 2peter = 0$$ for $$peter$$ in the interval $$0^\circ \leq peter \leq 180^\circ$$. 2. **Use the Pythagorean identity:** Recall that $$\sin^2 x = 1 - \cos^2 x$$. Substitute $$\sin^2 2peter$$: $$3(1 - \cos^2 2peter) + 8\cos 2peter = 0$$ 3. **Expand and simplify:** $$3 - 3\cos^2 2peter + 8\cos 2peter = 0$$ Rearranged: $$-3\cos^2 2peter + 8\cos 2peter + 3 = 0$$ Multiply both sides by $$-1$$ to simplify: $$3\cos^2 2peter - 8\cos 2peter - 3 = 0$$ 4. **Let $$x = \cos 2peter$$:** The quadratic equation becomes: $$3x^2 - 8x - 3 = 0$$ 5. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times (-3)}}{2 \times 3} = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm \sqrt{100}}{6}$$ $$x = \frac{8 \pm 10}{6}$$ 6. **Calculate the two roots:** - $$x_1 = \frac{8 + 10}{6} = \frac{18}{6} = 3$$ - $$x_2 = \frac{8 - 10}{6} = \frac{-2}{6} = -\frac{1}{3}$$ 7. **Check domain for cosine:** Since $$\cos 2peter$$ must be between $$-1$$ and $$1$$, $$x_1 = 3$$ is invalid. 8. **Solve for $$2peter$$:** $$\cos 2peter = -\frac{1}{3}$$ 9. **Find general solutions for $$2peter$$:** $$2peter = \cos^{-1} \left(-\frac{1}{3}\right)$$ or $$2peter = 360^\circ - \cos^{-1} \left(-\frac{1}{3}\right)$$ Calculate: $$\cos^{-1} \left(-\frac{1}{3}\right) \approx 109.47^\circ$$ So, $$2peter = 109.47^\circ$$ or $$2peter = 360^\circ - 109.47^\circ = 250.53^\circ$$ 10. **Divide by 2 to find $$peter$$:** $$peter = \frac{109.47^\circ}{2} = 54.74^\circ$$ $$peter = \frac{250.53^\circ}{2} = 125.27^\circ$$ 11. **Check if solutions are within the interval:** Both $$54.74^\circ$$ and $$125.27^\circ$$ are within $$0^\circ \leq peter \leq 180^\circ$$. **Final answer:** $$peter \approx 54.74^\circ, 125.27^\circ$$