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Whakaoti mō x
x=\frac{2\pi n_{1}}{3}+\frac{2\pi }{9}\text{, }n_{1}\in \mathrm{Z}
x=\frac{2\pi n_{1}}{3}+\frac{4\pi }{9}\text{, }n_{1}\in \mathrm{Z}
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5 raruraru e ōrite ana ki:
\cos ( 3x + \pi ) = 0.5
Ngā Raru Ōrite mai i te Rapu Tukutuku
How do you prove the reduction identities \displaystyle{\cos{{\left({x}+\pi\right)}}}=-{\cos{{\left({x}\right)}}} using a unit circle diagram?
https://socratic.org/questions/how-do-you-prove-the-reduction-identities-cos-x-pi-cos-x-using-a-unit-circle-dia
I am not sure this little diagram is enough but... Explanation: Have a look:
How to find all solutions of \cos (3x) + \cos (x) = 0 in [-\pi,\pi]
https://www.quora.com/How-do-I-find-all-solutions-of-cos-3x-+-cos-x-0-in-pi-pi
cos 3x+cos x=0 =>2 cos 2x cos x=0 =>cos 2x=0; or ; cos x=0 For x belong to[-pi,pi]; cos x=0 =>x = pi/2 , -pi/2 , 3 pi/2 , -3 pi/2 ...
How do you find the range of \displaystyle{f{{\left({x}\right)}}}={5}{\cos{{\left({x}+\pi\right)}}}+{3} ?
https://socratic.org/questions/how-do-you-find-the-range-of-f-x-5cos-x-pi-3
Alan P. Apr 15, 2015 \displaystyle{\cos{{\left({x}+\pi\right)}}} has a range of \displaystyle{\left[-{1},+{1}\right]} (because \displaystyle{\cos{{(}}} anything \displaystyle{)} ...
How to solve 0=\cos{2x}+\cos x
https://math.stackexchange.com/questions/859100/how-to-solve-0-cos2x-cos-x
remember the identity \cos (2x)=\cos^2 (x)-\sin^2(x)=2\cos ^2 (x)-1. You want to solve 2\cos ^2 (x)-1+\cos (x)=0. Let t= \cos (x). The question 2t^2+t-1=0 and is easier to solve. Can you take ...
How do you simplify \displaystyle{2}{\cos{{\left({10}{x}\right)}}}+{4}{\cos{{\left({5}{x}\right)}}} ?
https://socratic.org/questions/how-do-you-simplify-2cos-10x-4cos-5x
Simplify y = 2cos (10x) + 4cos (5x) Ans: (cos 5x - 1 - sqrt3)(cos 5x + 1 - sqrt3) Explanation: Replace 2cos (10x) by \displaystyle{4}{{\cos}^{{2}}{\left({5}{x}\right)}}-{2} , we get: y = 4cos^2 ...
How do you solve \displaystyle{\cos{{\left({x}\right)}}}{\left({\cos{{\left({x}\right)}}}-{1}\right)}={0} ?
https://socratic.org/questions/how-do-you-solve-cos-x-cos-x-1-0
\displaystyle{x}={0}+{2}{k}\pi\ \text{ or }\ {x}=\pi+{2}{k}\pi Explanation: \displaystyle{\left({\cos{{x}}}\right)}{\left({\cos{{x}}}\right)}-{1}={0}\ \text{ can be expressed as} \displaystyle{{\cos}^{{2}}{x}}={1} ...
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\cos ( 3x + \pi ) = 0.5
\sin ( x ) = 1
\sin ( x ) - cos ( x ) = 0
\sin ( x ) + 2 = 3
{ \tan ( x ) } ^ {2} = 4
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