SOLUTION: Hi,
The question I'm having trouble with is,
solve {{{ sin(2x)= sqrt(2)cos(x) }}}, for 0 < x < 2(pi)
I've tried expand and collecting to one side;
{{{ sin(x)cos(x)+ cos(x)si
Question 635686: Hi,
The question I'm having trouble with is,
solve , for 0 < x < 2(pi)
I've tried expand and collecting to one side;
but I've found that that taking cos(x) as a factor will only lead bead to the beginning.
Thankyou Found 2 solutions by jsmallt9, ewatrrr:Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website! , for 0 < x < 2(pi)
> I've tried expand and collecting to one side;
Good idea (although you could have left the 2sin(x)cos(x))
> but I've found that that taking cos(x) as a factor will only lead bead to the beginning.
?? Actually factoring out cos(x) is the right thing to do:
We now have a product that equals zero. (This is why we make one side of the equation zero.) Then we factor the other side and then use the Zero Product Property: or
Simplifying the second equation: or
Solving the second equation for sin(x): or or
Next we find the general solution. We should recognize that 0 is a special angle value for cos and that is also a special angle value for sin. So we will not be using our calculators.
For cos(x) = 0:
Only 0 (and angles co-terminal with zero) have a cos of 0. So:
For :
We should know that an angle whose sin is will have reference
angle of . We should also know that sin is positive in the 1st and 2nd quadrants. Putting these together we get: (for the 1st quadrant) (for the 2nd quadrant)
Simplifying the second equation we get: (for the 2nd quadrant)
All together, our general solution (all the x's that fit your equation) is:
Now the specific solution (the x's in the specified range). For this we replace the n's in the general solution equation with various integers until we are convinced we have found all the x's between 0 and . (Note: The range excludes 0 and . Only the numbers between them.)
From we get no x's that are between 0 and . When n is 0 or negative we get an x that is too low and when x is 1 or more we get an x that is too big.
From :
When n = 0,
When n = 1 or more, we get an x that is too big.
When n = -1 or below, we get an x that is too small.
From :
When n = 0,
When n = 1 or more, we get an x that is too big.
When n = -1 or below, we get an x that is too small.
