+0  
 
0
665
5
avatar+1832 

 Aug 29, 2014

Best Answer 

 #4
avatar+26226 
+10

Solution: B

$$\texttyle{2\sin^2{(x)}+sin(x)-1 = 0 \quad \cos{2x} = \cos^2{(x)}-\sin^2{(x)} \quad \cos^2{(x)} = 1 - \sin^2{(x)} }$$

$$\Rightarrow \sin{(x)} = \cos{(2x)}$$

 

 Sep 1, 2014
 #1
avatar+32822 
+10

First set y = sinx and solve:   2y2 + y - 1 = 0 for y.

 

Since the LHS factors nicely, we have (y+1)*(2y-1) = 0

so y = -1 and y = 1/2

 

sin(x) = -1 implies x = 3pi/2 + 2npi  where n is an integer

I'll leave you to work out what values of x make sin(x) = 1/2 .

 Aug 29, 2014
 #2
avatar+115747 
+10

Let y=sinx

 

Then solve for y

 

Then put the sinx back in and solve for x

 

Give it a go from there. 

 

Im on a little phone I will look again after im home :)

I see Alan beat me.  Thanks Alan :)

 Aug 29, 2014
 #3
avatar+1832 
+5

thank you alan and melody 

now its very clear 

 Aug 29, 2014
 #4
avatar+26226 
+10
Best Answer

Solution: B

$$\texttyle{2\sin^2{(x)}+sin(x)-1 = 0 \quad \cos{2x} = \cos^2{(x)}-\sin^2{(x)} \quad \cos^2{(x)} = 1 - \sin^2{(x)} }$$

$$\Rightarrow \sin{(x)} = \cos{(2x)}$$

 

heureka Sep 1, 2014
 #5
avatar+1832 
0

thank you heureka

 Sep 1, 2014

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