Question
Question: How do you solve \(\sin 2x-\cos x=0\)?...
How do you solve sin2x−cosx=0?
Solution
We will use the double angle formula for the sine function, which is given as sin2x=2sinxcosx. Then we will take the cosine function common. After that we will get two factors of the given equation such that their product is zero. Then we will equate these factors to zero and find the possible values of the variable x.
Complete step by step answer:
The given equation is sin2x−cosx=0. We know that there is a double angle formula for the sine function. This formula is given as sin2x=2sinxcosx. Substituting this formula in the given equation, we get the following expression,
2sinxcosx−cosx=0
Both the terms on the left hand side have cosx. We will take out cosx as a common factor. SO, we get the following equation,
cosx(2sinx−1)=0
Now, we have obtained two factors of the given expression such that their product is 0. We know that if the product of two numbers is 0, then either one of the two numbers has to be 0. Therefore, we have two possibilities, that is, either one of the factors in the above equation can be 0. So, we have either cosx=0 or 2sinx−1=0.
If cosx=0, then we know that x=2π. The cosine function is a periodic function with a period of 2π. Therefore, we have x=(2n+1)2π where n is a natural number.
If 2sinx−1=0, we have sinx=21. We know that if sinx=21, then x=6π. The sine function is a periodic function and its period is 2π. Therefore, we have x=6π+2πn where n is a natural number.
Therefore, the solution of the given equation is x=(2n+1)2π or x=6π+2πn where n is a natural number.
Note:
We should be familiar with the values of the trigonometric functions for standard angles. These values are useful in such types of questions. The double angle formulae and the half angle formulae are also essential in simplification of an equation. An important aspect of the trigonometric functions is their periodicity. A periodic function is a function that repeats its values after a fixed period.