Solve the equation 4sin^2 x = 3 in the range 0° < x < 360°
i worked out one of the answer is 60°
but the answer should be 60° , 240°
please explain
+36915 4 sin^2 = 3
sin^2 = 3/4
sin = +- (sqrt 3) /2 this occurs at 60 120 240 300 degrees
+2440 60° and 240° are not the only solutions in the interval 0 < x < 360. 120º and 300º are both perfectly valid.
Let's isolate the trigonometric term and see where to go from there.
| \(4\sin^2x=3\) | Divide by 4 on both sides. |
| \(\sin^2x=\frac{3}{4}\) | Take the square root of both sides. |
| \(|\sin x|=\sqrt{\frac{3}{4}}\) \(|\sin x|=\frac{\sqrt{3}}{2}\) | Use the definition of the absolute value to split this into two separate equations. |
| \(\sin x=\frac{\sqrt{3}}{2}\\ \sin x=\frac{-\sqrt{3}}{2}\) | |
Here, you locate which angle yields an answer of \(\pm\frac{\sqrt{3}}{2}\) , which is \(\{60^{\circ},120^{\circ},240^{\circ},300^{\circ}\}\)
+2440 Guest, I am not sure why the answer key would exclude the other valid solutions.
