What is 0<= 2x+pi/2 <= 2pi ?

The solution to 0<= 2x+pi/2 <= 2pi is -pi/4 <= x<= (3pi)/4

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0<= 2x+pi/2 <= 2pi

0 \leq 2 x + \frac{π}{2} \leq 2 π

- \frac{π}{4} \leq x \leq \frac{3 π}{4}

Solution steps

0 \leq 2 x + \frac{π}{2} \leq 2 π

a \leq u \leq b

then

a \leq u and u \leq b

0 \leq 2 x + \frac{π}{2} and 2 x + \frac{π}{2} \leq 2 π

0 \leq 2 x + \frac{π}{2} : x \geq - \frac{π}{4}

2 x + \frac{π}{2} \leq 2 π : x \leq \frac{3 π}{4}

Combine the intervals

x \geq - \frac{π}{4} and x \leq \frac{3 π}{4}

Merge Overlapping Intervals

- \frac{π}{4} \leq x \leq \frac{3 π}{4}

1 6^{2} + 1 2^{2}

n^{2} - 14 n + 48

\frac{2 - i}{3 + i} - \frac{1}{2 i}

(\frac{4 a b ^{2}}{5 ab})^{2}

\frac{- 2}{- 8}

What is 0<= 2x+pi/2 <= 2pi ?
The solution to 0<= 2x+pi/2 <= 2pi is -pi/4 <= x<= (3pi)/4