How do you solve for all solutions of x in $sin2x-1=0$ ?

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Answer 1 · 2015-02-15T01:39:05

(This is probably an attempt to try to trick you into using a double angle formula).

Let $theta = 2x$
So the problem becomes:
$sin(theta) - 1 = 0$
or
$sin(theta) = 1$

Within the range [ $0, 2Pi$ ]
$theta = Pi/2$ or $(3Pi)/2$
So (within the range [ $0, 2Pi$ ]
$x = Pi/4$ or $3Pi/4$

These solutions are repeated for every cycle of $2Pi$
so $x epsilon { Pi/4 + n2Pi, (3Pi)/4 + n2Pi }$
for all integer values of $n$

How do you solve for all solutions of x in sin2x-1=0sin2x-1=0?