Question

How do you solve ${\text{3si}}{{\text{n}}^{\text{2}}}{\text{x = co}}{{\text{s}}^{\text{2}}}{\text{x}}$ ?

Answer 1

Hint : We are given with an expression that has some trigonometric functions. We have to use some basic trigonometric identities to solve the question given above. We know that ${\text{si}}{{\text{n}}^2}x + {\cos ^2}x = 1$ . We will modify this identity to solve the expression above. On further solution we will get the value of x.

Complete step-by-step answer :
Given that,
${\text{3si}}{{\text{n}}^{\text{2}}}{\text{x = co}}{{\text{s}}^{\text{2}}}{\text{x}}$
So ${\cos ^2}x$ can also be written as ${\cos ^2}x = 1 - {\text{si}}{{\text{n}}^2}x$
So rewriting the given expression
$\Rightarrow {\text{3si}}{{\text{n}}^{\text{2}}}{\text{x}} = 1 - {\text{si}}{{\text{n}}^2}x$
Taking the sin terms on one side of the expression
$\Rightarrow {\text{3si}}{{\text{n}}^{\text{2}}}{\text{x + si}}{{\text{n}}^2}x = 1$
On adding both the terms
$\Rightarrow 4{\text{si}}{{\text{n}}^2}x = 1$
$4{\text{si}}{{\text{n}}^2}x$ can also be written in square form as $\Rightarrow {\left( {2{\text{sin}}x} \right)^2} = 1$
Now taking square root on both sides
$\Rightarrow 2{\text{sin}}x = 1$
So to find the value of x,
$\Rightarrow {\text{sin}}x = \dfrac{1}{2}$
$\Rightarrow x = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
We know that $\sin {30^ \circ } = \dfrac{1}{2}$ . So the value of x becomes,
$\Rightarrow x = {30^ \circ }$ or $\Rightarrow x = \dfrac{\pi }{6}$
This is the solution to the above problem.
So, the correct answer is “ $x = {30^ \circ }$ or $x = \dfrac{\pi }{6}$”.

Note : Note that when we solve this type of problem the question itself is the indicator to proceed. Here the cos function and sin function have relation in between. So that can be used to solve the same. Also note that ${\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}$ , ${\text{sin2x}}$ and ${\text{2sinx}}$ are totally different. In ${\text{si}}{{\text{n}}^{\text{2}}}{\text{x}}$ it is the square of the function, in ${\text{sin2x}}$ is it the angle is doubled and in ${\text{2sinx}}$ the function is doubled. Remember this. Their values are also different. When we make rearrangements in problems or substitute the functions this is really important. This is applicable for all trigonometric functions along with sin.