Question

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

Answer 1

Hint: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
\[\cos \left( {\alpha + \beta } \right) = \cos \left( \alpha \right)\cos \left( \beta \right) - \sin
\left( \alpha \right)\sin \left( \beta \right)\]
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.