Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 35

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 .