Solve for all real values of x
3cos2x - 5cosx = 1
Use cos2x = 2 cos^2x -1 and convert the equation to a second order polynomial in which cosx is the variable. There should be two possible values of cosx.
To solve the equation 3cos(2x) - 5cos(x) = 1 for all real values of x, we can use a trigonometric identity to simplify the equation.
First, let's use the double angle identity for cosine: cos(2x) = 2cos^2(x) - 1.
Now, our equation becomes:
3(2cos^2(x) - 1) - 5cos(x) = 1.
Expanding and rearranging terms, we obtain:
6cos^2(x) - 3 - 5cos(x) = 1.
Rearranging again, we have a quadratic equation:
6cos^2(x) - 5cos(x) - 4 = 0.
To solve this quadratic equation, we can factor or use the quadratic formula.
If you prefer factoring, you can look for two numbers whose product equals -24 (6 * -4), and whose sum equals -5. By inspection, we find that -8 and 3 satisfy these conditions. Therefore, we can split the middle term as follows:
6cos^2(x) - 8cos(x) + 3cos(x) - 4 = 0.
Re-arranging the equation, we have:
(6cos^2(x) - 8cos(x)) + (3cos(x) - 4) = 0,
Factoring by grouping, we get:
2cos(x)(3cos(x) - 4) + (3cos(x) - 4) = 0.
Now, we can factor the common term:
(2cos(x) + 1)(3cos(x) - 4) = 0.
Setting each factor equal to zero, we have two equations:
2cos(x) + 1 = 0 => 2cos(x) = -1 => cos(x) = -1/2,
3cos(x) - 4 = 0 => 3cos(x) = 4 => cos(x) = 4/3.
To find the solutions, we look for angles whose cosine is equal to -1/2 and 4/3. Using the unit circle or a trigonometric table, we find that for -1/2, the angles are 2pi/3 + 2πn and 4pi/3 + 2πn (where n is an integer), and for 4/3, the angle is arccos(4/3).
Therefore, the solutions for x are:
x = 2pi/3 + 2πn,
x = 4pi/3 + 2πn,
x = arccos(4/3) + 2πn (where n is an integer).
To solve the equation 3cos2x - 5cosx = 1 for all real values of x, we can use the double angle formula for cosine and simplify it.
The double angle formula for cosine is:
cos2x = 2cos²x - 1
Substituting this into the equation, we get:
3(2cos²x - 1) - 5cosx = 1
Expanding and rearranging the equation, we have:
6cos²x - 3 - 5cosx = 1
Rearranging again, we get:
6cos²x - 5cosx - 4 = 0
Now, let's solve this quadratic equation for cosx by factoring or using the quadratic formula.
We can factorize the equation as follows:
(2cosx - 1)(3cosx + 4) = 0
Setting each factor equal to zero, we have two possibilities:
2cosx - 1 = 0 or 3cosx + 4 = 0
Solving for cosx in the first equation:
2cosx = 1
cosx = 1/2
Solving for cosx in the second equation:
3cosx = -4
cosx = -4/3
So, there are two possible values for cosx: 1/2 and -4/3.
Now, we need to find the corresponding values of x.
Using the inverse cosine function, we can solve for x:
For cosx = 1/2:
x = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer.
For cosx = -4/3:
There is no solution for this case since the range of the cosine function is -1 to 1, and -4/3 is outside this range.