Question
Question: How do you solve \(\cos 2x + \sin x = 0\)?...
How do you solve cos2x+sinx=0?
Solution
Here we can turn the above-given equation into the quadratic equation in sinx and then we can find the general value of sinx for which it is satisfying and hence we can get the solution of the problem.
Complete step by step solution:
Here we need to find the value of x
So here we are given the equation as:
cos2x+sinx=0 −−−(1)
We know that cos2x=cos2x−sin2x −−−−−(2)
Also we know that
sin2x+cos2x=1 cos2x=1−sin2x
Now we can substitute this value in the equation (2) and get:
cos2x=1−sin2x−sin2x cos2x=1−2sin2x −−−−−(3)
Now substituting this value we get in equation (3) in the equation (1) we will get:
1−2sin2x+sinx=0 \-2sin2x+sinx+1=0
Now we get the quadratic equation in sinx as:
2sin2x−sinx−1=0
Now we can write in above equation that (−sinx)=(−2sinx+sinx)
We will get:
2sin2x−2sinx+sinx−1=0
Simplifying it further we will get:
2sinx(sinx−1)+(sinx−1)=0
(2sinx+1)(sinx−1)=0
So we can say either sinx=−21 or sinx=1
Now we can say that two value of the trigonometric function are −21,1
So value of x will be sin−1(21) or sin−1(1)
So this is just the general value of the x we have obtained but there can be many value of x as sinx=−21 or sinx=1 at many values of x
This is also because we are not given any particular interval in which we have to find the solution.
Note:
This problem can also be asked in the manner where we can be asked to find the value of x in the particular interval. Then we can find the particular values of x but here we need to give the answer as the general value which is sin−1(21) or sin−1(1).