(arcsin(x/2)+arccos(x/2))/arctan(x) = 3/2, solve for x if x >0. Thanks.
By definition, arcsin(x/2) would be the angle π, so that sinπ = x/2
Using a hunch that perhaps we would deal with one of our standard angles,
I tried x = 0, 1, β2, β3, 2 , (which would be sides in our standard 45-45-90 and 30-60-90 triangles
for x = 0 ,
(arcsin(0)+arccos(0))/arctan(0) = (0 + 1)/0 β 3/2
for x= 1
(arcsin(1/2)+arccos(1/2)/arctan(2) = (30 + 60)/63.43.. β 3/2
x = β2
(arcsin(β2/2)+arccos(β2/2)/arctan(β2) = (45+45)/54.7... β 3/2
x = β3
(arcsin(β3/2)+arccos(β3/2)/arctan(β3) = (60 + 30)/60 = 90/60 = 3/2
Well, how is that for a "lucky" guess
At the moment, I can't think of a formal way to solve this.
To solve the given equation, we will use some trigonometric identities and properties. Let's break it down step by step:
1. Let's start by using the property of the inverse trigonometric functions that states arcsin(x) + arccos(x) = Ο/2. By substituting x/2 into both arcsin and arccos functions, we have:
arcsin(x/2) + arccos(x/2) = Ο/2
2. Now we have the equation in the form:
(Ο/2) / arctan(x) = 3/2
3. To simplify further, we can take the reciprocal of both sides:
arctan(x) / (Ο/2) = 2/3
4. By using the property of the inverse tangent function that arctan(1/x) = arctan(x), we can rewrite the equation as:
arctan(1/x) / (Ο/2) = 2/3
5. Simplifying the equation further, we get:
2 * arctan(1/x) = (2/3) * (Ο/2)
arctan(1/x) = Ο/3
6. Now, we can take the tangent of both sides:
tan(arctan(1/x)) = tan(Ο/3)
1/x = β3
7. Taking the reciprocal of both sides, we find:
x = 1/β3
x = β3/3
Therefore, the solution to the equation (arcsin(x/2) + arccos(x/2))/arctan(x) = 3/2, with x > 0, is x = β3/3.