Question

Solve the following equation: $\left( \cos 2x-1 \right){{\cot }^{2}}x=-3\sin x$

Answer 1

Now to solve this question you will first convert the double angle into single angle by using the formulas for it which is $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$. Then on opening we can notice that we can convert the expression left can be made into a quadratic and solve it to find the value of trigonometric function and solve for the value of x.

Complete step-by-step solution:
Now the equation given to us here is ;
$\left( \cos 2x-1 \right){{\cot }^{2}}x=-3\sin x$
Now to simplify it we first start by simplifying the double function; now for that we know the formula that $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$, therefore using that we get
$\left( {{\cos }^{2}}x-{{\sin }^{2}}x-1 \right){{\cot }^{2}}x=-3\sin x$
Now through the identities of trigonometry which is ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ we can put the value of ${{\cos }^{2}}x$ in our equation and it gives us;
$\left( (1-{{\sin }^{2}}x)-{{\sin }^{2}}x-1 \right){{\cot }^{2}}x=-3\sin x$
Opening the bracket
$\left( 1-{{\sin }^{2}}x-{{\sin }^{2}}x-1 \right){{\cot }^{2}}x=-3\sin x$
Simplifying we get
$\left( -2{{\sin }^{2}}x \right){{\cot }^{2}}x=-3\sin x$
Now we know that we that the value of $\cot x=\dfrac{\cos x}{\sin x}$ ; therefore using that we can write the expression as
$\left( -2{{\sin }^{2}}x \right)\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}=-3\sin x$
Cancelling the same terms from both numerator and denominator
$-2{{\cos }^{2}}x=-3\sin x$
Now we use the trigonometry identity which is ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ to simplify this sum only in the form of sine functions;
$-2(1-{{\sin }^{2}}x)=-3\sin x$
Opening the bracket
$-2+2{{\sin }^{2}}x=-3\sin x$
Now since we can see that to solve this equation we can do it by simplifying it to a quadratic equation, that’s we here we can substitute the value of sine to be t; that is
$\sin x=t$
Therefore we get
$-2+2{{t}^{2}}=-3t$
We can write this as
$2{{t}^{2}}+3t-2=0$
Now here we can see that the sum of this quadratic equation is $3$ and the product is $-4$, therefore we can write it is
$2{{t}^{2}}+4t-t-2=0$
Taking $2t$ common from first two terms and $-1$ common from last two we get
$2t(t+2)-1(t+2)=0$
Therefore
$(2t-1)(t+2)=0$
So the values of t are
$t=\dfrac{1}{2};t=-2$
Now we need to find the value of sine so substituting
$\sin x=\dfrac{1}{2};\sin x=-2$
We know that the range of sine is from $\left[ -1,1 \right]$ therefore the value of sine can’t be $-2$
$\sin x=\dfrac{1}{2}$
Since sine is a periodic function it will give us the same value after the interval of $n\pi$ so we can write the answer of x as
$x=n\pi +{{(-1)}^{n}}\dfrac{\pi }{6}$

Note: A common mistake made by students here is just writing the answer of x to be $\dfrac{\pi }{6}$ . While that answer is definitely not wrong but since sine is a periodic function it doesn’t include all values of x and therefore despite $\dfrac{\pi }{6}$ also being a value of x it is not the only one. A common fact that students also need to remember is that the value of sine can’t be out of the range of $\left[ -1,1 \right]$. Any answer you get out of it is not possible.