Thread: Trig Equations

Need to solve solutions for [0,2π)

Problem: Sin x + cos X=1. What I did was square both side.
SinX^2+2sinXcosX+cosX^2=1. It's the Middle term that is really throwing me off. I know that I need to get either all cos or sin. I know how to change Cosx^2 into 1-sinx^2. Just not sure where do go after that.

Try this:



(Note: we may have just introduced spurious roots by this irreversible process. We have to check our final answers to make sure they work.)





You can use a trig identity on the LHS, I think. Does that help?

Yes, I would use sin2x. So, my final equation would look like sin2x=0?

Right. So what does x have to be?

Originally Posted by Ackbeet

Try this:



(Note: we may have just introduced spurious roots by this irreversible process. We have to check our final answers to make sure they work.)





You can use a trig identity on the LHS, I think. Does that help?

Of course, you don't really need to use that identity. If then either sin(x)= 0 or cos(x)= 0.

I think I get. If i divide both side by 2, that gives me sinx cosx = 0. So, sinx =0 and cosx =0. which would give the 3 solutions of: 0, pi/2, and 3pi/2.

Zero

Think of it this way: sin(2x) = 0 means that 2x must be equal to any of those points where the sin function is zero. That happens where? Keep in mind the original interval of interest.

O,Pi/2

I think there are two more, for a total of four. But the four solutions I'm thinking of have not yet been checked against the original equation.

Originally Posted by Brndo4u

Need to solve solutions for [0,2π)

Problem: Sin x + cos X=1. What I did was square both side.
SinX^2+2sinXcosX+cosX^2=1. It's the Middle term that is really throwing me off. I know that I need to get either all cos or sin. I know how to change Cosx^2 into 1-sinx^2. Just not sure where do go after that.

Here's also an alternative approach....

The maximum value of is 1. The maximum value of is also 1.

If you draw a Unit Circle, pick any point on the circumference,
the hypotenuse of any right-angled triangle (from the origin to a point on the circle) is 1 unit.

Hence, in the 1st quadrant, we cannot sum the perpendicular sides of the triangle to get a value of 1,
since the sum of the perpendicular sides exceeds the hypotenuse.

In the 2nd quadrant, we have is negative, but the maximum value of is 1,

thus the sum there.

We have a similar problem in quadrants 3 and 4,

hence we cannot have anywhere on the Unit Circle off the x and y axes.

Therefore, you only need to check the angles at the intersections of the axes.