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Algebraically, how can I find the exact solutions of the equation sqrt3 +2cos(2u)=0 in the interval [0,2pi)

Algebraically How Can I Find The Exact Solutions Of The Equation Sqrt3 2cos2u0 In The Interval 02pi class=

Sagot :

Solution:

Consider the following trigonometric equation:

[tex]\sqrt[]{3}+2\cos (2u)=0[/tex]

to solve this equation, subtract the root of 3 from both sides of the equation, to obtain:

[tex]2\cos (2u)=-\sqrt[]{3}[/tex]

now, solving for cos(2u), we get:

[tex]\cos (2u)=-\frac{\sqrt[]{3}}{2}[/tex]

now, by the trigonometric circle, the general solutions for the above equation are:

[tex]2u\text{ = }\frac{5\pi}{6}+2\pi n\text{ , 2u = }\frac{7\pi}{6}+2\pi n\text{ }[/tex]

now, if we solve the above equations for u, we get the following general solutions:

[tex]u\text{ = }\frac{5\pi}{12}+\pi n\text{ , u = }\frac{7\pi}{12}+\pi n\text{ }[/tex]

so, for the interval [0,2pi), the solution would be:

[tex]u\text{ = }\frac{5\pi}{12},\text{ u=}\frac{7\pi}{12},\text{ u=}\frac{17\pi}{12},\text{ u=}\frac{19\pi}{12}[/tex]

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