Solve for $x$ when $1-\sin^2x - \cos 2x = \dfrac{1}{2}$.
I can' t change it into a form I can work with. It is rather complicated.
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Assuming that "sinx^2" means $\sin(x)^2$, remember that $1-\sin(x)^2=\cos(x)^2 = \frac12(1+\cos(2x))$. You now have a linear equation in $\cos(2x)$. – hmakholm left over Monica May 24 '15 at 23:00
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Is that $\sin^{2}(x)$ or $\sin(x^2)$? – user222031 May 24 '15 at 23:00
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this is all one term as sin^2x. – user242559 May 24 '15 at 23:01
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@user242559 Please use appropriate formatting. See here: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Tim Raczkowski May 24 '15 at 23:07
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Hint: $\color{#0ae}{(1-\sin^2 x)}-\color{#0b4}{(\cos 2x)}=\color{#0ae}{(\cos^2 x)} -\color{#0b4}{(\cos^2x-\sin^2x)}=\sin^2x =\frac{1}{2}$
$\iff \sin x=\pm\frac{\sqrt{2}}{2}$. You should be able to solve this.
user26486
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2thanks for the first part of changing forms. now I've figured out the rest. – user242559 May 24 '15 at 23:14
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Alternate forms:
$sin^2(x)=\frac{1}{2}$
$\frac{1}{2}(1-cos(2x))=\frac{1}{2}$
$\frac{sin^2(x)}{2}-\frac{1}{2}cos^2(x)+\frac{1}{2}=\frac{1}{2}$
$\Longrightarrow$
$ x=\pi n-\frac{\pi}{4}, n \in \mathbb{Z}$
$x=\pi n+\frac{\pi}{4}, n \in \mathbb{Z}$
3SAT
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As given $$1-\sin^2 x-\cos 2x=\frac{1}{2} \implies \cos^2x-\left(2\cos^2 x-1\right)=\frac{1}{2}$$ $$ \cos^2 x=\frac{1}{2}=\cos^2 \frac{\pi}{4} \implies x=n\pi\pm \frac{\pi}{4}$$ Hence, the general solution of the equation is $x=n\pi\pm \frac{\pi}{4} $ where, $n$ is an integer.
Harish Chandra Rajpoot
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