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Popular >

cos^2(θ)0<θ<360

Solution

cos^{2} (θ) 0^{\circ} < θ < 36 0^{\circ}

0 < θ < 36 0^{\circ}

Interval Notation

(0, 36 0^{\circ})

Decimal Notation

0 < θ < 6.28318 \dots

Solution steps

cos^{2} (θ) \cdot 0^{\circ} < θ < 36 0^{\circ}

a < u < b

then

a < u and u < b

cos^{2} (θ) \cdot 0^{\circ} < θ and θ < 36 0^{\circ}

cos^{2} (θ) \cdot 0^{\circ} < θ : θ > 0

cos^{2} (θ) \cdot 0^{\circ} < θ

Apply rule

0^{a} = 0

0^{\circ} = 0

0 \cdot cos^{2} (θ) < θ

Apply rule

0 \cdot a = 0

0 < θ

Move

0

to the right side

0 < θ

Simplify

< θ

< θ

Move

θ

to the left side

< θ

Subtract

θ

from both sides

- θ < θ - θ

Simplify

- θ < 0

- θ < 0

Multiply both sides by

- 1

- θ < 0

Multiply both sides by -1 (reverse the inequality)

(- θ) (- 1) > 0 \cdot (- 1)

Simplify

θ > 0

θ > 0

Combine the intervals

θ > 0 and θ < 36 0^{\circ}

Merge Overlapping Intervals

θ > 0 and θ < 36 0^{\circ}

The intersection of two intervals is the set of numbers which are in both intervals

θ > 0

and

θ < 36 0^{\circ}

0 < θ < 36 0^{\circ}

0 < θ < 36 0^{\circ}

\frac{d}{d x} ((2 sin (x) - x) 0) \leq x \leq 2 π

sin (θ) cos (θ) = 0.222 and tan (θ) < 0

sin (θ) = 56, π^{2} \leq θ \leq π

arcsinh (\frac{8}{15})

\frac{( sin ( 56. 2 ^{\circ} ) \cdot 19 ) ^{2}}{( 2 \cdot 9.8 )}