Solve the equation for solutions in the interval 0<=theta<2pi Problem 1. 3cot^2-4csc=1 My attempt: 3(cos^2/sin^2)-4/sin=1 3(cos^2/sin^2) - 4sin/sin^2 = 1 3cos^2 -4sin =sin^2 3cos^2-(1-cos^2) =4sin 4cos^2 -1 =4sin Cos^2 -
Can anyone help w/ these. 1) Solve the equation in the internal [0deg, 360deg]. a) sin 2x = -sin x b) sin 2T = -1/2 (where T is angle) c) 4 sin^2T = 3 2) Evaluate the expression. sin(arctan 2) 3) Rewrite the following w/o using
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x +
Evaluate (exact answers): a) sin^-1(cos30) Need help. What do I do? b) sin[(cos^-1((sqrt2)/(2)))+[(sin^-1((sqrt2)/(2)))] = sin(45+45) = sin90 = 1 Is this correct? c) cos(arctan5/7) = 7/(sqrt74) = (7sqrt74)/74 Is this correct?
I would like to solve the ∫sin^2(pix) dx Using the given substitution identity: sin^2(x) = (1/2)(1-cos2x) This is what I did so far: ∫sin^2(pix) let u = pix du = pi dx (1/pi)∫sin^2(u)du Applying the identity is where I'm
I need to find the exact solutions on the interval [0,2pi) for: 2sin^2(x/2) - 3sin(x/2) + 1 = 0 I would start: (2sin(x/2)-1)(sin(x/2)-1) = 0 sin(x/2)=1/2 and sin(x/2)=1 what's next? Ok, what angle has a sin equal to say 1/2 sin
ok, i tried to do what you told me but i cant solve it for c because they cancel each others out! the integral for the first one i got is [sin(c)cos(x)-cos(c)sin(x)+sin(x)+c] and the integral for the 2nd one i got is