Question

The solution of the equation $4{{\cos }^{2}}x+6{{\sin }^{2}}x=5$
A. \[x=n\pi \pm \dfrac{\pi }{2}\]
B. \[x=n\pi \pm \dfrac{\pi }{4}\]
C. \[x=n\pi \pm \dfrac{3\pi }{2}\]
D. None of these.

Answer 1

Hint: we will rewrite the given equation in a way so that we can apply the formula of ${{\sin }^{2}}x+{{\cos }^{2}}x$ in it. We will then simplify the equation and derive an equation and then we will use the theorem according to which for any of the real numbers $x,y$, $\sin x=\sin y$ implies that \[x=n\pi \pm y\] where \[n\in Z\].

Formula Used: ${{\sin }^{2}}x+{{\cos }^{2}}x=1$

Complete step by step solution: We are given an equation $4{{\cos }^{2}}x+6{{\sin }^{2}}x=5$ and we have to determine the general solution of this equation.
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, so we will use in the given equation and simplify.
$\begin{align}
  & 4{{\cos }^{2}}x+6{{\sin }^{2}}x=5 \\
 & 4{{\cos }^{2}}x+4{{\sin }^{2}}x+2{{\sin }^{2}}x=5 \\
 & 4({{\sin }^{2}}x+{{\cos }^{2}}x)+2{{\sin }^{2}}x=5 \\
 & 4+2{{\sin }^{2}}x=5 \\
 & 2{{\sin }^{2}}x=1 \\
 & {{\sin }^{2}}x=\dfrac{1}{2} \\
 & \sin x=\pm \dfrac{1}{\sqrt{2}}
\end{align}$
Now we know that $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ and $\sin \left( -\dfrac{\pi }{4} \right)=-\dfrac{1}{\sqrt{2}}$so,
$\sin x=\sin \dfrac{\pi }{4}$ or $\sin x=\sin \left( -\dfrac{\pi }{4} \right)$
We will now apply the theorem here to get the solution of the equation.
$x=n\pi +\dfrac{\pi }{4}$ or $x=n\pi -\dfrac{\pi }{4}$
Therefore the solution will be \[x=n\pi \pm \dfrac{\pi }{4}\] where $n$is an integer.
The general solution of the trigonometric equation $\cos p\theta =\cos q\theta $ such that $p\ne q$,is $\theta =\dfrac{2n\pi }{p\pm q}$

Option ‘B’ is correct

Note: Instead of using this formula ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ we could have also used the formula ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ and substituted in the given equation. Both the formulas are almost similar and derive the same resultant equation.
The domain of the trigonometric function sin is set of all of the real numbers but its range is in between $\left[ -1,1 \right]$. In a right angled triangle, it can be defined as the ratio of perpendicular side of the triangle to the hypotenuse of the triangle.