Solve the equation sin^2x=3Cos^2X.
I would love to graduate...so thanks in advance!
![Solve the equation sin2x3Cos2X I would love to graduateso thanks in advance class=](https://us-static.z-dn.net/files/d2b/b57251902c2eeb9f57369b7e177c3804.png)
Answer:
240°
Step-by-step explanation:
[tex]sin^2x=3cos^2x \\ \\ sin^2x=3(1 - sin^2x )\\ \\ sin^2x=3 - 3sin^2x \\ \\ sin^2x + 3sin^2x=3 \\ \\ 4sin^2x=3 \\ \\ sin^2x= \frac{3}{4} \\ \\ sinx= \sqrt{ \frac{3}{4} } \\ \\ sinx= \frac{ \sqrt{3} }{2} \\ \\ sinx= sin \: 60 \degree \\ \\ in \: the \: second \: quadrant \\ sinx= sin \: (180 \degree - 60 \degree ) = sin \: 120 \degree \\ \\ in \: the \: third \: quadrant \\ sin x= sin \: (180 \degree + 60) \degree = sin \: 240 \degree[/tex]