Find the value of sinx ,if 8sinx-cosx=4
8sinx-cosx = 4
8sinx = cosx+4
64sin^2x = cos^2x + 8cosx + 16
64-64cos^2x = cos^2x + 8cosx + 16
65cos^2x + 8cosx - 48 = 0
(5cosx-4)(13cosx+12) = 0
cosx = 4/5 or -12/13
so, sinx = 3/5 or 5/13
To find the value of sin(x), we'll rearrange the given equation and solve for sin(x):
8sin(x) - cos(x) = 4
First, let's rearrange the equation:
8sin(x) = 4 + cos(x)
Now, we'll square both sides of the equation to eliminate the trigonometric functions:
(8sin(x))^2 = (4 + cos(x))^2
64sin^2(x) = 16 + 8cos(x) + cos^2(x)
Using the trigonometric identity sin^2(x) + cos^2(x) = 1:
64(1 - cos^2(x)) = 16 + 8cos(x) + cos^2(x)
64 - 64cos^2(x) = 16 + 8cos(x) + cos^2(x)
65cos^2(x) + 8cos(x) - 48 = 0
Now, we'll solve this quadratic equation for cos(x) using either factoring, completing the square, or the quadratic formula.
Let's solve for cos(x) using the quadratic formula:
cos(x) = (-8 ± √(8^2 - 4(65)(-48))) / (2(65))
cos(x) = (-8 ± √(64 + 12480)) / 130
cos(x) = (-8 ± √12544) / 130
cos(x) = (-8 ± 112) / 130
cos(x) = (104 / 130) or (-120 / 130)
cos(x) = 8/10 or -12/10
cos(x) = 4/5 or -6/5
Now, to find sin(x), we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1:
sin^2(x) + (4/5)^2 = 1 or sin^2(x) + (-6/5)^2 = 1
sin^2(x) + 16/25 = 1 or sin^2(x) + 36/25 = 1
sin^2(x) = 9/25 or sin^2(x) = -11/25 (not possible in real numbers)
Taking the square root of both sides, we have:
sin(x) = ± (√9/√25)
sin(x) = ± (3/5)
Therefore, the values of sin(x) are either 3/5 or -3/5.