Please help I need to solve the following question for x, where 0<x<2pi. and then write a general solution for the equation. The question is 2cos^2 x + cos x =1. I need to solve this using radian measurement. I solved it using algebra and was told I wasn't allowed. Could you please show me the steps to solve it using radian measurement and then how to write a general equation also using radian measurement. Please and thanks
hey Rae,
there is two ways you can do this...
1/
find the common factor of 2cos^2x+cosx=1
and bring out the front which = cosx(2cosx+1)=1
therefore either cosx=1 OR 2cosx+1=1 (because there will be multiple answers)
for cosx=1 use the unit circle to show where cosx=1 which is on the x axis because cos(pi)=1 which is show at points 0 and pi on the unit circle.
for 2cosx+1=1 simplify to cosx=-1/2
cos(pi/3)=-1/2 therefore x=pi/3
graph this in the 2nd and 3rd quadrant (s and t) because cos is negative in those quadrants
you should now have 4 "lines" on your unit circle graph. start at zero going anti clock wise labelling the points.
which will be x=0, (pi- pi/3), pi, (pi+pi/3) which simplified will equal
0, 2pi/3, pi, 4pi/3 these are your four possible answers for x.
2/ the other way to do this is using the quadratic formula.
in the equation 2cos^2x+cosx=1 , let cosx=t and substitute that in
so your equation should now be 2t^2+t=1
take the 1 over to the other side so it is 2t^2+t-1=0 which gives you a quadratic put this into the quadratic formula
so it will be now t= 1 +or- the square root (1^2-4*2*-1)all divided by 2*2
which simplifies to t=(1+or- 3)/4
therefore t= 1 OR -1/2
sub cosx=t into that
therefore cosx=1 OR cosx=-1/2
its the the same as 1/ from now on..
hope this helps.:D
I disagree with "belle"
2cos^2 x + cos x =1
2cos^2 x + cos x -1 = 0
(2cosx - 1)(cosx + 1) = 0
cosx = 1/2 or cosx = -1
case1: cosx = 1/2
we know cosπ/6 = 1/2 and the cosine is positive in I and IV
so x = π/6 or x = 2π - π/6 = 11π/6
case 2: cosx = -1
x = π
so x = π/6 , π, 11π/6
since the period of cosx is 2π, general solutions would be
π/6 + 2kπ
π + 2kπ or π(1+2k)
11π/6 + 2kπ, where k is an integer.
Sure! I can guide you through solving the equation 2cos^2 x + cos x = 1 using radian measurement. Here are the steps:
Step 1: Rewrite the equation in terms of cosine.
The equation is already in terms of cosine. So, we can move on to the next step.
Step 2: Rearrange the equation to form a quadratic equation.
2cos^2 x + cos x - 1 = 0
Step 3: Let cosine x be a variable.
Let's substitute cos x with another variable, say u, to simplify the equation:
2u^2 + u - 1 = 0
Step 4: Solve the quadratic equation for u.
You can solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
If we solve this equation using the quadratic formula, we have:
u = (-b ± √(b^2 - 4ac)) / (2a)
a = 2, b = 1, c = -1
u = (-1 ± √(1^2 - 4(2)(-1))) / (2(2))
u = (-1 ± √(1 + 8)) / 4
u = (-1 ± √9) / 4
u = (-1 ± 3) / 4
So, we have two possible values for u:
u1 = (3 - 1) / 4 = 2 / 4 = 1/2
u2 = (-1 - 3) / 4 = -4 / 4 = -1
Step 5: Replace the variable u with cos x.
Since 0 < x < 2π, we know that the cosine function will have positive values. Therefore, -1 is not a valid solution for cos x.
So, cos x = 1/2
Step 6: Find the solutions for x.
To find the solutions for x, we need to locate the angles where cos x equals 1/2 within the given range (0 < x < 2π). These angles can be found using the inverse cosine function (also known as arccos or cos⁻¹).
cos⁻¹(1/2) = π/3
So, x can be:
x = π/3, or
x = 2π - π/3 = 5π/3
These are the specific solutions for x within the given range.
Step 7: Write the general solution (using radian measurement).
To write the general solution, we need to consider the periodicity of the cosine function. The cosine function repeats every 2π radians. So, we can add or subtract a multiple of 2π from any specific solution to obtain more solutions.
In this case, the general solution for x is:
x = π/3 + 2kπ, or
x = 5π/3 + 2kπ
where k is an integer.
These equations give us the general solution for x within the given range (0 < x < 2π).
I hope this explanation helps! Let me know if you have any further questions.