Question

Solve for exact solutions over [0,2pie) interval.
sin^2 x/2-2=0

Answer 1

1) Solve the equation:

$$\sin^ {2} \frac {x}{2} - 2 = 0$$

Solution:

$$\sin^ {2} \frac {x}{2} - 2 = 0$$

Add 2 to both sides

$$\sin^ {2} \frac {x}{2} = 2$$

Take the square root of both sides

$$\sin \frac {x}{2} = \sqrt {2} \quad \text {or} \quad \sin \frac {x}{2} = - \sqrt {2}$$

Look at the first equation $\sin \frac{x}{2} = \sqrt{2}$ .

Take the inverse sine of both sides:

$$\frac {x}{2} = \pi - \arcsin \sqrt {2} + 2 \pi n _ {1} n _ {1} \in Z$$

or

$$\frac {x}{2} = \arcsin \sqrt {2} + 2 \pi n _ {2} n _ {2} \in Z$$

Multiply both sides by 2:

$$x = 2 \pi - 2 \arcsin \sqrt {2} + 2 \pi n _ {1} n _ {1} \in Z$$ $$x = 2 \arcsin \sqrt {2} + 4 \pi n _ {2} n _ {2} \in Z$$

Look at the second equation $\sin \frac{x}{2} = -\sqrt{2}$

Take the inverse sine of both sides:

$$\frac {x}{2} = \pi + \arcsin \sqrt {2} + 2 \pi n _ {3} n _ {3} \in Z$$ $$\frac {x}{2} = 2 \pi n _ {4} - \arcsin \sqrt {2} n _ {4} \in Z$$

Multiply both sides by 2:

$$x = 2 \pi + 2 \arcsin \sqrt {2} + 4 \pi n _ {3} n _ {3} \in Z$$ $$x = 4 \pi n _ {4} - 2 \arcsin \sqrt {2} n _ {4} \in Z$$

Answer:

$$x = 2 \pi - 2 \arcsin \sqrt {2} + 2 \pi n _ {1} n _ {1} \in Z$$ $$x = 2 \arcsin \sqrt {2} + 4 \pi n _ {2} n _ {2} \in Z$$ $$x = 2 \pi + 2 \arcsin \sqrt {2} + 4 \pi n _ {3} n _ {3} \in Z$$ $$x = 4 \pi n _ {4} - 2 \arcsin \sqrt {2} n _ {4} \in Z$$

2) Solve for exact solutions over $[0, 2\mathrm{pie})$ interval:

$$\sin^ {2} \left(\frac {x}{2} - 2\right) = 0$$

Solution:

Take the square root of both sides:

$$\sin \left(\frac {x}{2} - 2\right) = 0$$

Take the inverse sine of both sides:

$$\frac {x}{2} - 2 = \pi n n \in Z$$ $$0 \leq x < 2 \pi$$

so

$$\frac {x}{2} - 2 = 0$$ $$\frac {x}{2} - 2 = \pi$$

Add 2 to both sides:

$$\frac {x}{2} = 2$$ $$\frac {x}{2} = \pi + 2$$

Multiply both sides by 2:

$$x = 4$$ $$x = 2 \pi + 4$$

Answer:

$$x = 4$$ $$x = 2 \pi + 4$$