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

Guest Feb 3, 2019

#1**0 **

4 sin^2 = 3

sin^2 = 3/4

sin = +- (sqrt 3) /2 this occurs at 60 120 240 300 degrees

ElectricPavlov Feb 3, 2019

#2**+1 **

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 \(\{60^{\circ},120^{\circ},240^{\circ},300^{\circ}\}\)

.TheXSquaredFactor Feb 3, 2019

#4**+1 **

Guest, I am not sure why the answer key would exclude the other valid solutions.

TheXSquaredFactor
Feb 3, 2019