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Popular >

cos^2(x)-2cos(x)+1<= 0

Solution

cos^{2} (x) - 2 cos (x) + 1 \leq 0

Solution

False for all x \in R

Solution steps

cos^{2} (x) - 2 cos (x) + 1 \leq 0

Let:

u = cos (x)

u^{2} - 2 u + 1 \leq 0

u^{2} - 2 u + 1 \leq 0 : u = 1

u^{2} - 2 u + 1 \leq 0

Factor

u^{2} - 2 u + 1 : (u - 1)^{2}

u^{2} - 2 u + 1

Rewrite

u^{2} - 2 u + 1

as

u^{2} - 2 u \cdot 1 + 1^{2}

u^{2} - 2 u + 1

Rewrite

1

as

1^{2}

= u^{2} - 2 u + 1^{2}

Rewrite

2 u

as

2 u \cdot 1

= u^{2} - 2 u \cdot 1 + 1^{2}

= u^{2} - 2 u \cdot 1 + 1^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = u, b = 1

= (u - 1)^{2}

(u - 1)^{2} \leq 0

For

u^{n} \leq 0

, if

n

is even then

u = 0

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

u = 1

Substitute back

u = cos (x)

cos (x) = 1

General solutions for

cos (x) = 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = 0 + 2 πn

x = 0 + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

No Solution for x \in R

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