Sketch the following equations over an interval 0 2pi 2sin2

Sketch the following equations over an interval [0, 2pi]. 2sin^2 x - six x - 1 = 0 Answer: cos x + 1 = 2sin^2x. Answer: Sketch the graph of the function f(x) = - 1/2 cos (2x + pi) and find the amplitude and period. Do not use a calculator.

Solution

4) 2sin^2x - sinx -1 =0

2sin^2x -2sinx +sinx -1 =0

2sin(sinx -1)+1(sinx -1) =0

(2sinx+1)(sinx-1) =0

sinx = -1/2 ---> x = pi+pi/6 , 2pi -pi/6

x = 7pi/6 , 11pi/6

sinx -1 =0

sinx =1---> x =pi/2

Solution : x = pi/2, 7pi/6 , 11pi/6

5) cosx +1 = 2sin^2x

substitute sin^2x = 1-cos^2x

cosx +1 = 2(1-cos^2x) = 2 -2cos^2x

2cos^2x +cosx -1 =0

factorise: 2cos^2x +2cosx - cosx -1 =0

2cosx(cosx +1) -1(cosx+1) =0

(2cosx -1)(cosx +1) =0

cosx =1/2 ---x = pi/3, 2pi -pi/3

In the interval [0, 2pi] x = pi/3 , 5pi/3

cosx +1 =0

cosx=-1----> x= pi

Solution : In the interval [0, 2pi] x= pi/3 , pi ,5pi/3