+0

# Solve the equation 4sin^2 x = 3 in the range 0 < x < 360

0
264
4

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°

Feb 3, 2019

#1
+19811
0

4 sin^2 = 3

sin^2 = 3/4

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

Feb 3, 2019
#2
+2344
+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 $$\{60^{\circ},120^{\circ},240^{\circ},300^{\circ}\}$$

.
Feb 3, 2019
edited by TheXSquaredFactor  Feb 3, 2019
#3
+1

i'm sorry but my answer sheet states the only 2 answers should be 60 and 240

Guest Feb 3, 2019
#4
+2344
+1

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

TheXSquaredFactor  Feb 3, 2019