Question

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

Answer 1

Hint: 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.