Solve for x using inverse trig identities

Question

arcsin(sqrt(2x))=arccos(sqrt(x))

Answer 1

arcsin (√2x) → sin θ = √2x

arccos (√x) → cos θ = √x, These are by definition that should be learned.

sin² θ + cos² θ = 1, Pythagorean Identity

2x + x = 1

x = ?

Answer 2

arcsin(sqr2x) = arccos(sqrx)

the angle whose sine = cosine is 45 degrees or pi/4 radians

as cos45 =sin45 = sqr2/2=about .707, but

the angle whose sine =sqr2x = the angle whose cosine = sqrx = the angle whose tangent = sqr2

which = arctan(sqr2)

the angle = about 1 radian = about 54.74 degrees= about .955 radians

it's an angle with height=sqr2, base=1, hypotenuse = sqr3

x = 1/3

Answer 3

I've seen a fair amount of students who were otherwise doing well in pre-calc and calc AB/BC suddenly struggle when it comes to solving inverse trig related questions. Since these are usually lectured on in their own topic, these A or B students would suddenly start failing tests and experiencing a lot of anxiety. I created this to hopefully help those of you who got by with a passing knowledge of inverse trig to be more prepared if/when your classes focus on these types of functions.

Answer 4

To solve the equation arcsin⁡(2x)=arccos⁡(x),

we can use the fact that

arcsin⁡(y)+arccos⁡(y)=π2 for any y in the interval [0,1]

Thus, we can rewrite the equation as:

arcsin⁡(2x)+arccos⁡(x)=π2

Substituting arccos⁡(x) from the original equation gives:

arcsin⁡(2x)+arcsin⁡(2x)=π2

This simplifies to:

arcsin⁡(2x)=π2−arccos⁡(x)

Given that arcsin⁡(y)+arccos⁡(y)=π2, we can find that:

2x=cos⁡(arccos⁡(x))=x

Squaring both sides:

2x=x

Rearranging this gives:

2x−x=0 ⟹ x=0

Now we need to check if there are any other solutions. To explore, we can substitute back into the original functions:

1. Check the domain: 2x and x​ must be valid inputs for arcsin⁡ and arccos⁡, respectively. Hence, x must be in the interval [0,1/2] for 2x​ and [0,1] for x​.
2. Plugging x=0 into the original equation:
3. arcsin(0)=arccos(0) which holds true since both equal 0.

Thus the solution is 0.