Hint:
You'll use very similar steps found here
https://socratic.org/questions/how-do-you-solve-arcsin-x-arcsin-2x-pi-3
Though instead of
Caution: The link posted has an error in one of the steps. Specifically the step where it says this
The
Let "a" be the angle -<= a <= , sin(a) = x, and Let "b" be the angle - <= b <= , sin(b) = 2x. They want you find the value of x such that b - a = . (1) Since sin(a) = x and sin(b) = 2x, we have cos(a) = , cos(b) = . From equation (1), taking cosine from both sides, we have this equation cos(b-a) = , or cos(b)*cos(a) + sin(b)*sin(a) = , or, substituting + (2x)*x = , or = - 2x^2. Now square both sides. You will get then (1-4x^2)*(1-x^2) = - 2x^2 + 4x^4. Simplify it step by step 1 - 4x^2 - x^2 + 4x^4 = - 2x^2 + 4x^4 1 - 5x^2 = - 2x^2 4 - 20x^2 = 1 - 8x^2 4 - 1 = 20x^2 - 8x^2 3 = 12x^2 1 = 4x^2 x^2 = x = = +/- . Thus the equation is just solved, and we have two potential solutions x= +/- . Consider these two cases separately and check the results in both cases. Case a). x = ; arcsin(x) = ; 2x = 1; arcsin(2x) = arcsin(1) = . Since - = = = , the solution is correct. Case b). x = - ; arcsin(x) = - ; 2x = -1; arcsin(2x) = arcsin(-1) = - . Since - = = - = - , this solution DOES NOT work. It is EXTRANEOUS. +--------------------------------------------------------------------+ | So, the problem has a unique solution | | | | x = , and arcsin(x) = , arcsin(2x) = . | +--------------------------------------------------------------------+
1) The solution under the link
https://socratic.org/questions/how-do-you-solve-arcsin-x-arcsin-2x-pi-3
mentioned by tutor @Math_tutor2020, is INCORRECT.
2) The solution by @Math_tutor2020 also contains a technical error.
It is WHY I came to bring the correct and accurate solution.