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Question: How do you solve \(\cos x-3\cos \left( \dfrac{x}{2} \right)=0\) ?...

How do you solve cos⁡x−3cos⁡(x2)=0\cos x-3\cos \left( \dfrac{x}{2} \right)=0 ?

Explanation

Solution

We know the formula cos 2x is equal to 2cos⁡2x−12{{\cos }^{2}}x-1 from this formula we can prove that cos x is equal to 2cos⁡2x2−12{{\cos }^{2}}\dfrac{x}{2}-1. Now we can use this formula to solve the given question. We can replace cos x with 2cos⁡2x2−12{{\cos }^{2}}\dfrac{x}{2}-1 , then the equation will convert into a quadratic equation where the variable is cos⁡(x2)\cos \left( \dfrac{x}{2} \right) .

Complete step-by-step answer:
We have to solve cos⁡x−3cos⁡(x2)=0\cos x-3\cos \left( \dfrac{x}{2} \right)=0
We can write cos x is equal to 2cos⁡2x2−12{{\cos }^{2}}\dfrac{x}{2}-1. Now if we replace cos x with 2cos⁡2x2−12{{\cos }^{2}}\dfrac{x}{2}-1. It will change into a quadratic equation.
2cos⁡2x2−1−3cos⁡(x2)=02{{\cos }^{2}}\dfrac{x}{2}-1-3\cos \left( \dfrac{x}{2} \right)=0
We can see that the above equation is a quadratic equation where the variable is cos⁡(x2)\cos \left( \dfrac{x}{2} \right)
The formula roots of any quadratic equation is −b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} where a is the coefficient of x2{{x}^{2}} , b is coefficient of x and c is the constant in a quadratic equation.
In the equation 2cos⁡2x2−1−3cos⁡(x2)=02{{\cos }^{2}}\dfrac{x}{2}-1-3\cos \left( \dfrac{x}{2} \right)=0 , a = 2 , b = -3 and c = -1
Substituting the value of a, b and c in the formula, the roots are
−3±(−3)2−4×2×−12×2=−3±174\dfrac{-3\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 2\times -1}}{2\times 2}=\dfrac{-3\pm \sqrt{17}}{4}
So the roots are −3+174\dfrac{-3+\sqrt{17}}{4} and −3−174\dfrac{-3-\sqrt{17}}{4}
We know that the range of cos x is from – 1 to 1. So −3−174\dfrac{-3-\sqrt{17}}{4} can not be the root of the equation. The only possible root is −3+174\dfrac{-3+\sqrt{17}}{4}
So the value of cos⁡(x2)\cos \left( \dfrac{x}{2} \right) is equal to −3+174\dfrac{-3+\sqrt{17}}{4}
So x is equal to 2cos⁡−1−3+1742{{\cos }^{-1}}\dfrac{-3+\sqrt{17}}{4} and the general solution is 2nπ+2cos⁡−1−3+1742n\pi +2{{\cos }^{-1}}\dfrac{-3+\sqrt{17}}{4} where n is an integer.

Note: While solving any polynomial equation where the variable is any trigonometric function or exponential function, always check the root whether it lies in the range of that function. For example if the variable is an exponential function, then negative roots will be invalid.