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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

 Feb 3, 2019
 #1
avatar+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
avatar+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
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+1

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

Guest Feb 3, 2019
 #4
avatar+2344 
+1

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

TheXSquaredFactor  Feb 3, 2019

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