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
Hello, vecoma!
How do i solve this equation?
. . $\displaystyle \sin(2x) \:=\: -\dfrac{\sqrt{2}}{2}$
$\displaystyle \text{We have: }\:\sin(2x) \:=\:-\dfrac{1}{\sqrt{2}}$
From your basic trig values, you should know that:
. . the angle whose sine is $\displaystyle \text{-}\frac{1}{\sqrt{2}}\,\text{ is }\,\dfrac{5\pi}{4}\,\text{ or }\,\dfrac{7\pi}{4}$
$\displaystyle \text{So we have: }\;2x \;=\;\begin{Bmatrix}\dfrac{5\pi}{4} + 2\pi n \\ \\[-3mm] \dfrac{7\pi}{4} + 2\pi n \end{Bmatrix}$
. . $\displaystyle \text{Therefore: }\;x \;=\;\begin{Bmatrix}\dfrac{5\pi}{8} + \pi n \\ \\[-3mm] \dfrac{7\pi}{8} + \pi n \end{Bmatrix}\;\text{ for some integer }n$
Hi vecoma,
$\displaystyle sin2x=2sinxcosx$ 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 $\displaystyle sinA=-\frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{\sqrt{2}\;\sqrt{2}}$
where $\displaystyle A=2x$
first find A, then x