How do I find the value of x given sinx = 1 + cos²x?

1 Answer
Feb 25, 2018

x=pi/2+2pin where n is any integer.

Explanation:

Recall the identity

sin^2(x)+cos^2(x)=1

Rewriting a bit, we get

cos^2(x)=1-sin^2(x)

So, we can rewrite our equation in terms of only sin(x):

sin(x)=1+(1-sin^2(x))

sin(x)=2-sin^2(x)

Let's move everything to the left side and have the right side be 0:

sin^2(x)+sin(x)-2=0

This looks like a quadratic equation, and we can treat it as one. The difference is, instead of having x, we have sin(x).

Let's factor this just as we would factor any other quadratic:

(sin(x)+2)(sin(x)-1)=0

Solving, we get

sin(x)=-2

sin(x)=1

Now, we need to solve each of the above equations for x:

sin(x)=-2

There is no solution to this. The range of sin(x) is -1<=x<=1, looking either at the graph or unit circle tells us this, as the function oscillates forever between -1 and 1 . sin(x) can never be >1 or <-1.

sin(x)=1

Let's consider where sin(x) is equal to 1.

We know sin(pi/2)=1, so x=pi/2 is certainly a valid answer. However, x=(5pi)/2 is also a valid answer, as (5pi)/2 means one and a half counterclockwise rotations around the unit circle, which would cause us to end up again at pi/2. The same applies for (9pi)/2, two and a half counterclockwise rotations around the unit circle, or x=-(3pi)/2, one and a half clockwise rotations around the unit circle.

Basically, adding or subtracting 2pi (or any multiple of 2pi) to pi/2 gives us a value which causes sin(x) to equal 1. This is due to the periodic nature of sin(x). The period of sin(x) is 2pi, so the value for sin(x) at a certain value of x repeats every 2pi units .

So, we can denote our answer by

x=pi/2+2pin where n is any integer, thereby accounting for every possible answer.