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

To solve, let's consider the right side of the equation first. Let it be equal to `theta`.

`theta = arccos(sqrtx)`

Then, express it in terms of cosine.

`cos (theta)= sqrtx`

Also, express the `sqrtx` as a fraction.

`cos(theta)=sqrtx/1`

Based on the formula `cos (theta)= (adjacent)/(h y p o t e n u s e)` , it can be deduced that the two sides of the right triangle are:

adjacent side =`sqrt x`

hypotenuse = 1`

To solve for the expression that represents the side opposite the theta, apply Pythagorean formula.

`a^2+b^2=1`

`a^2+(sqrtx)^2=1^2`

`a^2+x=1`

`a^2=1-x`

`a=+-sqrt(1-x)`

Since a represents the length of the opposite side, consider only the positive expression. So the opposite side is

opposite side`= sqrt(1-x)`

Now that the expression that represents the three sides of the triangle are known, let's consider the original equation again.

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

Plug-in the assumption that `theta = arccos(sqrt(x))` .

`arcsin (sqrt(2x)) = theta`

Then, express the equation in terms of sine function.

`sqrt(2x)=sin(theta)`

To express the right side in terms of x variable, refer to the right triangle. Applying the formula `sin (theta) = (opposite)/(hypotenuse)` , the right side becomes

`sqrt(2x) = (sqrt(1-x))/1`

`sqrt(2x)=sqrt(1-x)`

Then, eliminate the square root in the equation.

`(sqrt(2x))^2 =(sqrt(1-x))^2`

`2x = 1 - x`

Bring together the terms with x on one side of the equation.

`2x + x = 1 -x + x`

`3x = 1`

And, isolate the x.

`(3x)/3=1/3`

`x=1/3`

**Therefore, the solution is `x=1/3` .**

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