Question

How do you solve $\arcsin \left( x \right) + \arcsin \left( {2x} \right) = \dfrac{\pi }{3}$ ?

Answer 1

To solve the given equation apply various trigonometric functions and find out the general quadratic equation and then apply its formula to get the value of $x$.

Complete step by step answer:
The given equation is
$\arcsin \left( x \right) + \arcsin \left( {2x} \right) = \dfrac{\pi }{3}$
Let us consider the given equation as
$\alpha + \beta = \dfrac{\pi }{3}$
In which $\alpha$ and $\beta$ are represented as given angles of the equation as:
$\alpha = \arcsin \left( x \right)$
$\beta = \arcsin \left( {2x} \right)$
Let us consider
$\sin \left( \alpha \right) = x$
As we know that $\sin \left( \alpha \right) = \sqrt {1 - {{\sin }^2}\left( \alpha \right)}$, so let us apply to the equation as
$\cos \left( \alpha \right) = \sqrt {1 - {{\sin }^2}\left( \alpha \right)}$
Which implies,
$\cos \left( \alpha \right) = \sqrt {1 - {x^2}}$ …………………… 1
Now, let us consider
$\sin \left( \beta \right) = 2x$
In the same way $\sin \left( \beta \right) = \sqrt {1 - {{\sin }^2}\left( \beta \right)}$, so let us apply to the equation as
$\cos \left( \beta \right) = \sqrt {1 - {{\sin }^2}\left( \beta \right)}$
Which implies,
$\cos \left( \beta \right) = \sqrt {1 - \left( {2{x^2}} \right)}$
$\cos \left( \beta \right) = \sqrt {1 - 4{x^2}}$ ………………….. 2
Next, consider
$\alpha + \beta = \dfrac{\pi }{3}$
Which implies
$\cos \left( {\alpha + \beta } \right) = \cos \left( {\dfrac{\pi }{3}} \right)$
As we know the formula to expand $\cos \left( {\alpha + \beta } \right)$is

\left( \alpha \right)\sin \left( \beta \right)$$ Applying the formula, we get $$\cos \left( \alpha \right)\cos \left( \beta \right) - \sin \left( \alpha \right)\sin \left( \beta \right) = \dfrac{1}{2}$$ ……………….. 3 Substitute the values of equation 1 and equation 2 in equation 3 we get $$\sqrt {1 - {x^2}} \cdot \sqrt {1 - 4{x^2}} - \left( x \right) \cdot \left( {2x} \right) = \dfrac{1}{2}$$ Combining all the x terms we get $$\sqrt {1 - {x^2} - 4{x^2} - 4{x^4}} = 2{x^2} + \dfrac{1}{2}$$ Squaring on both the sides of the equation as $${\left[ {\sqrt {1 - {x^2} - 4{x^2} - 4{x^4}} } \right]^2} = {\left[ {2{x^2} + \dfrac{1}{2}} \right]^2}$$ Simplifying the terms, we get $$1 - 5{x^2} - 4{x^4} = 4{x^4} + 2{x^2} + \dfrac{1}{4}$$ $$8{x^4} + 7{x^2} - \dfrac{3}{4} = 0$$ Hence, we get $$32{x^4} + 28{x^2} - 3 = 0$$ As the obtained equation is quadratic equation, hence let us apply the formula given as $$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$ Applying the quadratic formula in the variable $${x^2}$$i.e., $${x^2} = \dfrac{{ - 28 \pm \sqrt {{{28}^2} - 4\left( {32} \right)\left( { - 3} \right)} }}{{2\left( {32} \right)}}$$ $${x^2} = \dfrac{{ - 28 \pm \sqrt {784 + 384} }}{{64}}$$ $${x^2} = \dfrac{{ - 28 \pm \sqrt {1168} }}{{64}}$$ $${x^2} = \dfrac{{ - 28 \pm \sqrt {16.73} }}{{64}}$$ $${x^2} = \dfrac{{ - 7 \pm \sqrt {73} }}{{16}}$$ **Therefore, for the given equation the value of $$x$$ is true for $$x = \sqrt {\dfrac{{ - 7 + \sqrt {73} }}{{16}}} $$** **Additional information:** The number $${b^2} - 4ac$$ is called "discriminant". If D < 0, then the quadratic equation has no real solutions (it has 2 complex solutions), if D = 0, then the quadratic equation has 1 solution and If D > 0, then the quadratic equation has 2 distinct solutions. **Formula used:** Every quadratic equation can have 0, 1 or 2 real solutions derived by the formula: $$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$ where a, b, c are real numbers. The number $${b^2} - 4ac$$ is called "discriminant ". **Note:** The key point to find the value of $$x$$ in the given equation of the type $$\arcsin \left( x \right) + \arcsin \left( {2x} \right) = \dfrac{\pi }{3}$$, just apply various trigonometric functions to solve the equations and further simplifying the terms with respect to obtained equation.