solve for x if 2cos(x-30)=1, £(-90;270)
don't know where to start
recall:
cos(A-B) = cosAcosB + sinAsinB
so ....
2cos(x-30) = 1
2(cosxcos30 + sinxsin30) = 1
cosx(√3/2) + sinx (1/2) = 1/2
times 3
√3cosx + sinx = 1
√3cosx = 1 - sinx
square both sides
3cos^2 x = 1 - 2sinx + sin^2 x
3(1 - sin^2 x) = 1 - 2sinx + sin^2 x
...
4sin^2 x - 2sinx -2 = 0
2sin^2 x - sinx - 1 = 0
(2sinx + 1)(sinx - 1) = 0
sinx = -1/2 or sinx = 1
x = 210° or 330° or x = 90°
since we squared, all answers should be tested in the original question
if x = 210,
LS = 2cos(180) = -2 ≠ 1
if x = 330,
LS = 2cos300 = 1 = RS, so x = 330°
if x = 90
LS = 2cos(60) = 1 = RS , so x = 90°
x = 90° or x = 330°
in radians:
x = π/2, 11π/6
To solve for x in the equation 2cos(x - 30) = 1, we need to isolate the variable x.
1. Start by dividing both sides of the equation by 2:
(2cos(x - 30))/2 = 1/2
This simplifies to: cos(x - 30) = 1/2.
2. Next, we need to find the angle whose cosine is 1/2. You can do this by recalling the values of cosine for specific angles. In this case, cos(x - 30) = 1/2 is true for certain angles.
The unit circle provides us with the angles where the cosine is positive and equal to 1/2. These angles are 60 degrees (π/3 radians) and 300 degrees (5π/3 radians).
3. In the given interval, £(-90; 270), we need to find the values of x that satisfy the equation cos(x - 30) = 1/2.
Starting from -90 degrees (-π/2 radians) to 270 degrees (3π/2 radians), we can add 30 degrees (π/6 radians) to each potential angle to determine if it satisfies the equation.
-90° + 30° = -60° (-π/3 radians) --> cos(-60°) = 1/2 (satisfies the equation)
-60° + 30° = -30° (π/6 radians) --> cos(-30°) = √3/2 (does not satisfy the equation)
-30° + 30° = 0° (0 radians) --> cos(0°) = 1 (does not satisfy the equation)
...
Continue this process and you will find that x = 150° (5π/6 radians) and x = 390° (13π/6 radians) satisfy the equation.
Therefore, x = 150° and x = 390° are the solutions to the equation 2cos(x - 30) = 1 within the given interval £(-90; 270).