Question

How do you solve $\arcsin \left( {\sqrt {2x} } \right) = \arccos \left( {\sqrt x } \right)$ ?

Answer 1

Hint : To find the given trigonometric function, apply trigonometric identity formulas for cosec, as we know that sec, csc and cot are derived from primary functions of sin, cos and tan, hence using these functions we need to solve using trigonometric identities functions and hence by this we can get the value of $x$ .

Complete step-by-step answer :
Let us write the given equation:
$\arcsin \left( {\sqrt {2x} } \right) = \arccos \left( {\sqrt x } \right)$
The given equation is rewritten as:
$\sin \left( {\arccos \left( {\sqrt x } \right)} \right) = \sqrt {2x}$ ………………… 1
Square on both sides of the equation 1 we get:
$\Rightarrow {\sin ^2}\left( {\arccos \left( {\sqrt x } \right)} \right) = \left| {2x} \right|$ …………………. 2
Use trigonometric identities to convert the sine function to cosine function i.e., we know that ${\sin ^2}x = 1 - {\cos ^2}x$ , hence applying this to the equation 2 we get:
$\Rightarrow 1 - {\cos ^2}\left( {\arccos \left( {\sqrt x } \right)} \right) = \left| {2x} \right|$
Now, rearrange the terms of the equation, we get:
$\Rightarrow 1 - \left| {2x} \right| = {\cos ^2}\left( {\arccos \left( {\sqrt x } \right)} \right)$
Now, apply square root on both the sides of the obtained equation as:
$\sqrt {1 - \left| {2x} \right|} = \sqrt {{{\cos }^2}\left( {\arccos \left( {\sqrt x } \right)} \right)}$
$\Rightarrow \sqrt {1 - \left| {2x} \right|} = \cos \left( {\arccos \left( {\sqrt x } \right)} \right)$
Now,
$\sqrt {1 - \left| {2x} \right|} = \sqrt x$
Evaluating the square root terms as:
$\Rightarrow 1 - \left| {2x} \right| = x$
Hence, we get:
$\Rightarrow x = - 1$ or $x = \dfrac{1}{3}$
Therefore, $x = \dfrac{1}{3}$ , as x cannot be a negative value.
So, the correct answer is “$x = \dfrac{1}{3}$”.

Note : In trigonometry sin, cos and tan values are the primary functions we consider while solving trigonometric problems. These trigonometry values are used to measure the angles and sides of a right-angle triangle. Apart from sine, cosine and tangent values, other values are cotangent, secant and cosecant.
The trigonometric values are about the knowledge of standard angles for a given triangle as per the trigonometric ratios. Trigonometric ratios are Sine, Cosine, Tangent, Cotangent, Secant and Cosecant.
The key point to find the values of any trigonometric function is to note the chart of all functions as shown and calculate all the terms asked. And here are some of the formulas and integration terms to be noted.
${\sin ^2}x = 1 - {\cos ^2}x$ , ${\sin ^2}\theta + {\cos ^2}\theta = 1$ ,
$\int {\sin xdx = - \cos x + c}$ , $\int {\cos xdx = \sin x + c}$ , $\int {{e^x}dx = {e^x} + c}$