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Popular >

cos^2(x)>= sin^2(x)

Solution

cos^{2} (x) \geq sin^{2} (x)

Solution

- \frac{π}{4} + πn \leq x \leq \frac{π}{4} + πn

Interval Notation

[- \frac{π}{4} + πn, \frac{π}{4} + πn]

Decimal Notation

- 0.78539 \dots + πn \leq x \leq 0.78539 \dots + πn

Solution steps

cos^{2} (x) \geq sin^{2} (x)

Move

sin^{2} (x)

to the left side

cos^{2} (x) \geq sin^{2} (x)

Subtract

sin^{2} (x)

from both sides

cos^{2} (x) - sin^{2} (x) \geq sin^{2} (x) - sin^{2} (x)

cos^{2} (x) - sin^{2} (x) \geq 0

cos^{2} (x) - sin^{2} (x) \geq 0

Use the following identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

cos (2 x) \geq 0

For

cos (x) \geq a

, if

- 1 < a < 1

then

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (0) + 2 πn \leq 2 x \leq arccos (0) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

- arccos (0) + 2 πn \leq 2 x and 2 x \leq arccos (0) + 2 πn

- arccos (0) + 2 πn \leq 2 x : x \geq - \frac{π}{4} + πn

- arccos (0) + 2 πn \leq 2 x

Switch sides

2 x \geq - arccos (0) + 2 πn

Simplify

- arccos (0) + 2 πn : - \frac{π}{2} + 2 πn

- arccos (0) + 2 πn

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2} + 2 πn

2 x \geq - \frac{π}{2} + 2 πn

Divide both sides by

2

2 x \geq - \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{2 x}{2} \geq - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} \geq - \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

- \frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : - \frac{π}{4} + πn

- \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= - \frac{π}{4} + πn

x \geq - \frac{π}{4} + πn

x \geq - \frac{π}{4} + πn

x \geq - \frac{π}{4} + πn

2 x \leq arccos (0) + 2 πn : x \leq \frac{π}{4} + πn

2 x \leq arccos (0) + 2 πn

Simplify

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2} + 2 πn

2 x \leq \frac{π}{2} + 2 πn

Divide both sides by

2

2 x \leq \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{2 x}{2} \leq \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} \leq \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x \leq \frac{π}{4} + πn

x \leq \frac{π}{4} + πn

x \leq \frac{π}{4} + πn

Combine the intervals

x \geq - \frac{π}{4} + πn and x \leq \frac{π}{4} + πn

Merge Overlapping Intervals

- \frac{π}{4} + πn \leq x \leq \frac{π}{4} + πn

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sin(a)-cos(a)>0

sin (a) - cos (a) > 0

sin(pi/4-x)>= (sqrt(2))/2

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