Solve Interval of X Homework Equation

[PhysicsKid703]

Homework Statement

2sin²x + 3sinx - 2>0 and x² - x - 2 < 0 . x lies in the interval

A)∏/6 to 5∏6
B)-1 to 5∏/6
C)-1 to 2
D) ∏/6 to 2

Correct Answer is D, and I get C.

Homework Equations

The Attempt at a Solution

So I factorized the second equation and obtained (x+1)(x-2)<0
So this, on drawing a number line, shows that x lies between -1 and 2, both excluded.

Next, I factorized the first equation, which gave me (2sinx-1)(sinx+2)>0
Plotting on a number line, sin x is either less than -2 or greater than 1/2. Former is not possible.
That indicates that x lies between ∏/6 (=approx 0.523) and 5∏/6 (= approx 2.61), and all periodic repetitions of those two angles, as sin x is only greater than 1/2 for those values in between.

So the intersection of the first equation's number line( from -1 to 2) and this equations number line( from 0.523 to 2.61), gives me -1 to 2. This is apparently wrong, according to the answer in the book, which is ∏/6 to 2.

I'm rather tired, so I can't wrap my head around any stupid mistakes. Please do help :)

 

[LCKurtz]

Homework Statement

2sin²x + 3sinx - 2>0 and x² - x - 2 < 0 . x lies in the interval

A)∏/6 to 5∏6
B)-1 to 5∏/6
C)-1 to 2
D) ∏/6 to 2

Correct Answer is D, and I get C.

Homework Equations

The Attempt at a Solution

So I factorized the second equation and obtained (x+1)(x-2)<0
So this, on drawing a number line, shows that x lies between -1 and 2, both excluded.

Correct so far.

Next, I factorized the first equation, which gave me (2sinx-1)(sinx+2)>0
Plotting on a number line, sin x is either less than -2 or greater than 1/2. Former is not possible.
That indicates that x lies between ∏/6 (=approx 0.523) and 5∏/6 (= approx 2.61), and all periodic repetitions of those two angles, as sin x is only greater than 1/2 for those values in between.

Also correct.

So the intersection of the first equation's number line( from -1 to 2) and this equations number line( from 0.523 to 2.61), gives me -1 to 2.

No, that gives you $(\frac \pi 6, 2)$. The values of $x$ must be in both solution sets so anything less than $\frac \pi 6$ or greater than $2$ doesn't work.

 

[PhysicsKid703]

Wow. That was dumb of me.
Thanks so much.