2sinx-cosx=3/sqrt. of 2
2sinx = 3/√2 + cosx
4sin^2 x = 9/2 + 3√2 cosx + cos^2 x
4 - 4cos^2 x = 9/2 + 3√2 cosx + cos^2 x
5cos^2 x + 3√2 cosx + 1/2 = 0
cosx = [-3√2 ± √(18-10)]/10
= (-3√2 ± 2√2)/10
= -√2/10 or -1/√2
now just take arccos of those values.
Obviously x = 3π/4 is one, but I dunno the other.
To find the value of x in the equation 2sin(x) - cos(x) = 3/√2, we can use algebraic techniques to simplify and solve for x. Here's how:
Step 1: Square both sides of the equation to eliminate the square root:
(2sin(x) - cos(x))^2 = (3/√2)^2
Squaring the left side:
(2sin(x))^2 - 2(2sin(x))(cos(x)) + (cos(x))^2 = 3^2/2
Simplifying:
4sin^2(x) - 4sin(x)cos(x) + cos^2(x) = 9/2
Step 2: Use the trigonometric identity sin^2(x) + cos^2(x) = 1 to simplify the equation:
4(1 - cos^2(x)) - 4sin(x)cos(x) + cos^2(x) = 9/2
Distribute and combine like terms:
4 - 4cos^2(x) - 4sin(x)cos(x) + cos^2(x) = 9/2
Simplifying:
5cos^2(x) - 4sin(x)cos(x) - 4 = 9/2
Multiplying both sides by 2 to eliminate the fraction:
10cos^2(x) - 8sin(x)cos(x) - 8 = 9
Rearranging terms:
10cos^2(x) - 8sin(x)cos(x) - 17 = 0
Step 3: Use the double-angle identity for cos(2x) = 2cos^2(x) - 1:
5(2cos^2(x) - 1) - 8(sin(x))(cos(x)) - 17 = 0
Simplifying:
10cos^2(x) - 5 - 16sin(x)cos(x) - 17 = 0
Combining like terms:
10cos^2(x) - 16sin(x)cos(x) - 22 = 0
Step 4: Use the identity sin(2x) = 2sin(x)cos(x) to simplify:
10cos^2(x) - 16(sin(2x))/2 - 22 = 0
Simplifying:
10cos^2(x) - 8sin(2x) - 22 = 0
Step 5: Use the identity cos^2(x) = 1 - sin^2(x) to further simplify:
10(1-sin^2(x)) - 8sin(2x) - 22 = 0
Expanding:
10 - 10sin^2(x) - 8sin(2x) - 22 = 0
Combining like terms:
-10sin^2(x) - 8sin(2x) - 12 = 0
Step 6: Factor out a common factor:
2(-5sin^2(x) - 4sin(2x) - 6) = 0
Step 7: Solve for sin(x) using the quadratic formula:
-5sin^2(x) - 4sin(2x) - 6 = 0
Let's denote sin(x) as t:
-5t^2 - 4(2t)(√1-t^2) - 6 = 0
Simplifying:
-5t^2 - 8t√1-t^2 - 6 = 0
This equation can be solved using the quadratic formula. The resulting values of t can then be used to determine the corresponding values of x.