Hi...again.
I know I'm a bother but can you guys help me again?
I'm having trouble with these ones. Thanks!
1. find the exact solutions algebraically in the interval [0, 2pi)
sin 2x sin x = cos x. (2 is not an exponent on this one)
2. verify identity algebraically
cos^4 x - sin^4 x = cos 2x.
3. find all solutions of the equation in the interval [0, 2pi)
sin(x + pi/6) - sin(x - pi/6) = 1/2
sin(2x)*sin(x) = cos(x)
2*sin(x)*cos(x)*sin(x) = cos(x)
2*sin^2(x)*cos(x) - cos(x) = 0
cos(x)*[2*sin^2(x) - 1] = 0
Thus
cos(x) = 0 ==> x = (pi)/2, 3(pi)/2
or
2*sin^2(x) - 1 = 0
2*sin^2(x) = 1
sin^2(x) = 1/2
sin(x) = (+/-)1/sqrt{2}
x = (pi)/4, 3(pi)/4, 5(pi)/4, 7(pi)/4
Thus
x = (pi)/4, (pi)/2, 3(pi)/4, 5(pi)/4, 3(pi)/2, 7(pi)/4
-Dan