# Math Help - Trig Equations

1. ## 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.

2. Try this:

$\sin(x)+\cos(x)=1$

$\sin^{2}(x)+2\sin(x)\cos(x)+\cos^{2}(x)=1$ (Note: we may have just introduced spurious roots by this irreversible process. We have to check our final answers to make sure they work.)

$2\sin(x)\cos(x)+1=1$

$2\sin(x)\cos(x)=0.$

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

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

4. Right. So what does x have to be?

5. Originally Posted by Ackbeet
Try this:

$\sin(x)+\cos(x)=1$

$\sin^{2}(x)+2\sin(x)\cos(x)+\cos^{2}(x)=1$ (Note: we may have just introduced spurious roots by this irreversible process. We have to check our final answers to make sure they work.)

$2\sin(x)\cos(x)+1=1$

$2\sin(x)\cos(x)=0.$

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 $2\sin(x)\cos(x)= 0$ then either sin(x)= 0 or cos(x)= 0.

6. 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.

7. Zero

8. 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.

9. O,Pi/2

10. 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.

11. 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 $Cosx$ is 1. The maximum value of $Sinx$ 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 $Cosx$ is negative, but the maximum value of $Sinx$ is 1,

thus the sum $Sinx+Cosx<1$ there.

We have a similar problem in quadrants 3 and 4,

hence we cannot have $Sinx+Cosx=1$ 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.