Question

The equation $3{\sin ^2}x + 10\cos x - 6 = 0$ is satisfied if
A.$x = n\pi \pm {\cos ^{ - 1}}\left( {\frac{1}{3}} \right)$
B.$x = 2n\pi \pm {\cos ^{ - 1}}\left( {\frac{1}{3}} \right)$
C.$x = n\pi \pm {\cos ^{ - 1}}\left( {\frac{1}{6}} \right)$
D.$x = 2n\pi \pm {\cos ^{ - 1}}\left( {\frac{1}{6}} \right)$

Answer 1

Hint: We are given an equation and are asked to find the value of x which will satisfy the equation. Firstly let's frame the equation with all terms in terms of cos x so using \[{\sin ^2}x = 1 - {\cos ^2}x\] we get a quadratic equation in cos x. Let's solve the quadratic equation by splitting the middle term and find the appropriate value of x

Complete step-by-step answer:
Step 1 :
We are given that $3{\sin ^2}x + 10\cos x - 6 = 0$…………..(1)
We are asked to find for what value of x will the equation be satisfied.
Hence it is enough we find the value of by solving the above quadratic equation
Firstly, lets convert the equation in terms of cosx
We know that, ${\cos ^2}x + {\sin ^2}x = 1$
From this we get \[{\sin ^2}x = 1 - {\cos ^2}x\]……………(2)
Lets substitute this in the equation (1)
$
   \Rightarrow 3(1 - {\cos ^2}x) + 10\cos x - 6 = 0 \\
   \Rightarrow 3 - 3{\cos ^2}x + 10\cos x - 6 = 0 \\
   \Rightarrow - 3{\cos ^2}x + 10\cos x - 3 = 0 \\
   \Rightarrow 3{\cos ^2}x - 10\cos x + 3 = 0 \\
$
Now we have a quadratic equation in cosx
Step 2
Now let's solve the equation by factorisation method.
The middle term of the equation $ - 10\cos x$ can be split into two parts $ - 9\cos x - \cos x$
$ \Rightarrow 3{\cos ^2}x - \cos x - 9\cos x + 3 = 0$
Now let's take cosx common from the first two terms and and 3 common from the next two terms we get,
$
   \Rightarrow \cos x(3\cos x - 1) - 3(3\cos x - 1) = 0 \\
   \Rightarrow (3\cos x - 1)(\cos x - 3) = 0 \\
    \\
$
From this we get the value of cosx to be $\frac{1}{3}$ , 3
cosx = 3 is not possible
So cos x = $\frac{1}{3}$
From this $x = 2n\pi \pm {\cos ^{ - 1}}(\frac{1}{3})$
Therefore $x = 2n\pi \pm {\cos ^{ - 1}}(\frac{1}{3})$ satisfies the equation $3{\sin ^2}x + 10\cos x - 6 = 0$

The correct option is B

Note: We can solve the quadratic equation even by using the formula method.
That is , for a quadratic equation$a{x^2} + bx + c = 0$the roots are given by
$x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here , $a = 3,b = - 10,c = 3$
$
   \Rightarrow \cos x = \frac{{ - ( - 10) \pm \sqrt {{{( - 10)}^2} - 4(3)(3)} }}{{2(3)}} \\
   \Rightarrow \cos x = \frac{{10 \pm \sqrt {100 - 36} }}{6} \\
   \Rightarrow \cos x = \frac{{10 \pm \sqrt {64} }}{6} \\
   \Rightarrow \cos x = \frac{{10 \pm 8}}{6} \\
   \Rightarrow \cos x = \frac{{18}}{6},\frac{2}{6} \\
   \Rightarrow \cos x = 3,\frac{1}{3} \\
$
Hence it's up to the student to select the method to solve the quadratic equation.