Question
Question: How do you solve \[\sin 2x + \sin x = 0\] over the interval \[0\] to \[2\pi \]?...
How do you solve sin2x+sinx=0 over the interval 0 to 2π?
Solution
In this question, we have the trigonometric function. Which must solve for the interval0to2π. In the left-hand side, a trigonometric function sin2x will be there. To solve the above trigonometric equation, we use the formula. And the formula is given as below.
⇒sin2a=2sinacosa
Complete step by step answer:
In this question, an equation of xis given, which I want to solve. First we know about an equation, an equation is defined as it has two things which are equal. And the equation also likes a statement “this equal that”. The equation has two things or two sides, the left side is known as the left hand side and the right side is known as right hand side. The left hand side is denoted as “LHS” and the right hand side is denoted as “RHS”.
Now, come to the question. The equation is given below.
⇒sin2x+sinx=0
First, we take the left hand side from the above equation and want to solve it.
Then, the left hand side is.
sin2x+sinx
We know that, sin2a=2sina.cosa
Then above the left hand side is written as below.
⇒sin2x+sinx=2sinxcosx+sinx
Now we write right hand side is equal to right hand side.
Then,
2sinx.cosx+sinx=0
We solve the above equation as below.
We take the sinx as common from the left hand side.
Then,
sinx(2cosx+1)=0
We know that, if the product of any number of terms is equal to zero then one of the terms must equal to zero. Then
Now, we find the value of xfor the interval0to2π.
Then,
sinx=0
For the interval0to2π, the value ofxis.
x=0,x=πAndx=2π
2cosx+1=0
$$
2\cos x = - 1 \\
\cos x = - \dfrac{1}{2} \\