Let's solve for cosx using the quadratic formula:
Given equation: 6cos² x – 5cosx + 1 = 0
Let a = 6, b = -5, and c = 1.
cosx = [ -(-5) ± √((-5)² - 4(6)(1)) ] / 2(6)
cosx = [ 5 ± √(25 - 24) ] / 12
cosx = [ 5 ± √1 ] / 12
cosx = [ 5 ± 1 ] / 12
Now, we have two possible solutions for cosx:
Therefore, the solutions for cosx are cosx = 0.5 and cosx = 0.3333.
Let's solve for cosx using the quadratic formula:
Given equation: 6cos² x – 5cosx + 1 = 0
Let a = 6, b = -5, and c = 1.
cosx = [ -(-5) ± √((-5)² - 4(6)(1)) ] / 2(6)
cosx = [ 5 ± √(25 - 24) ] / 12
cosx = [ 5 ± √1 ] / 12
cosx = [ 5 ± 1 ] / 12
Now, we have two possible solutions for cosx:
cosx = (5 + 1) / 12 = 6 / 12 = 0.5cosx = (5 - 1) / 12 = 4 / 12 = 0.3333Therefore, the solutions for cosx are cosx = 0.5 and cosx = 0.3333.