Question

Find the exact solution(s) to the equation 2 sin²x-cosx + 1 = 0

Answer 1

Answer: x=2πn (n=0, 1, 2, 3,...)

Explanation:

2sin²x-cosx+1=0

2(1-cos²x)-cosx+1=0

2(1)-2(cos²x)-cosx+1=0

2-2cos²x-cosx+1=0

-2cos²x-cosx+3=0

Multiply both parts of the equation by -1:

2cos²x+cosx-3=0

Let cosx=t

Domain:

-1≤cosx≤1

Hence, -1≤t≤1

2t²+t-3=0

D=b²-4ac

a=2 b=1 c=-3

D=1²-4(2)(-3)

D=1+24

D=25

√D=√25

√D=5

`\displaystyle\\t=(-bб√(D) )/(2a) \\\\t=(-1б5)/(2(2))\\\\t=(-1б5)/(4)\\\\t=-1.5\\otin domain\\\\t=1\in domain\\\\cosx=1\\\\x=2\pi n\ where\ n=0, \ 1,\ 2,\ 3,\ ...`