+0

1 question !

0
665
5 Aug 29, 2014

#4
+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
+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
+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
+5

thank you alan and melody

now its very clear

Aug 29, 2014
#4
+10 Solution: B

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

$$\Rightarrow \sin{(x)} = \cos{(2x)}$$ heureka Sep 1, 2014
#5
0

thank you heureka

Sep 1, 2014