Question
Question: How do you solve \[2co{{s}^{2}}x+cosx=0\] and find all exact general solutions?...
How do you solve 2cos2x+cosx=0 and find all exact general solutions?
Solution
For solving the given expression take cosx common from the expression given in the question, after taking common equalize and compare the equations to determine the general solution of cosx. As, general solution for cosx=0 is given by;
x=2(2n+1)π, apply this to solve the expression given in the question.
Formula used:
-
The general solution for cosx=0 will be,
x=2(2n+1)π, where n is integer.
This is because the cosxhas a value equal to 0 at 2π,23π,25π,−27π,−211π, etc. -
The general solution of the trigonometric functions involved equations is given by,
x=nπ±θ, where n is an integer.
Complete step by step solution:
We have given that,
2cos2x+cosx=0
Taking cosx as common, we will get
∴cosx(2cosx+1)=0
Hence, from above expression,
We can say that,
Either the first part will be equal to zero or the second part will be equal to zero.
So,
Either cosx=0 or (2cosx+1)=0
If cosx=0
And
If (2cosx+1)=0
Then,
⇒cosx=2−1
Hence, general solution for cosx=0will be,
x=2(2n+1)π, where n is integer.
And
The general solution for cosx=2−1=cos(±32π) is x=nπ±32π, where n is an integer.
Hence General solution for 2cos2x+cosx=0 will be,
x=2(2n+1)π or x=2π±32π, where n is an integer.
Note:
While solving the equations involving trigonometric functions and asking for to find the general equations, students should need to remember the general solutions of all the trigonometric functions involved equations.
Here in the given question;
We have,
∴cosx(2cosx+1)=0
Hence, from above expression we can conclude that, either the first part will be equal to zero or the second part will be equal to zero.
So, the general equation of the expression given in the question will be an addition of the general equation of both the values.