How do i solve this equation?

sin(2x) = -(sqrt(2))/2

x = ___ ?

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i think i have to use: sin(2x) = 2sinxcosx

2. $\displaystyle x=\frac{sin^{-1}\left(\frac{-\sqrt{2}}{2}\right)}{2}+2\pi k \ \k\in\mathbb{Z}$

3. Hello, vecoma!

How do i solve this equation?

. . $\sin(2x) \:=\: -\dfrac{\sqrt{2}}{2}$

$\text{We have: }\:\sin(2x) \:=\:-\dfrac{1}{\sqrt{2}}$

From your basic trig values, you should know that:

. . the angle whose sine is $\text{-}\frac{1}{\sqrt{2}}\,\text{ is }\,\dfrac{5\pi}{4}\,\text{ or }\,\dfrac{7\pi}{4}$

$\text{So we have: }\;2x \;=\;\begin{Bmatrix}\dfrac{5\pi}{4} + 2\pi n \\ \\[-3mm] \dfrac{7\pi}{4} + 2\pi n \end{Bmatrix}$

. . $\text{Therefore: }\;x \;=\;\begin{Bmatrix}\dfrac{5\pi}{8} + \pi n \\ \\[-3mm] \dfrac{7\pi}{8} + \pi n \end{Bmatrix}\;\text{ for some integer }n$

4. Originally Posted by vecoma
How do i solve this equation?

sin(2x) = -(sqrt(2))/2

x = ___ ?

------------------

i think i have to use: sin(2x) = 2sinxcosx
Hi vecoma,

$sin2x=2sinxcosx$ expresses the sine of a double angle in terms of the single angle.

It only introduces complexity to your problem as you now have 2 trigonometric terms instead of 1.

Instead $sinA=-\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{\sqrt{2}\;\sqrt{2}}$

where $A=2x$

first find A, then x