Could someone help me with those two questions. I need to know how to do them without calculator
solve for the interval [0,2pie]
1) cos^2*x-cos2x=0
2) tan(x/2)-sinx=0
1.
cos^sup2;(x)-cos(2x) = 0
(1-sin²(x)) - (cos²(x)-sin²(x)) = 0
(1-sin²(x)) - cos²(x) + sin²(x) = 0
1-cos²(x) = 0
sin²(x) = 0
On the interval [0,2π],
sin²(x) = 0 at x=0, π and 2π
Subsititute x=0, π and 2π into the original equation to check each of the solutions.
2.
a. 2 trivial solutions
By inspection, at x=0 and x=2π,
tan(x/2)=sin(x)=0
b. other solutions
use
tan(x/2)=(1-cos(x))/sin(x), and
1-sin²(x) = cos²(x)
to simplify the expression to:
cos(x)(cos(x)-1)/sin(x) = 0
or cos(x)=0, or cos(x)=1
to give x=π/2, 3π/2, 0, 2π
Check each solution to confirm its validity.