Solve for x which is in [0,2π] 6cos⁡x+22sin⁡x=22 I have solved the question by dividing both sides 44, and got the answer that involves arcsin function. My question is: Is it possible to solve it without any arc functions?

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Solve for x which is in [0,2π]  6cos⁡x+22sin⁡x=22  I have solved the question by dividing both sides 44, and got the answer that involves arcsin function. My question is:  Is it possible to solve it without any arc functions?

usaho4w · Accepted Answer

6cos⁡x=22−22sin⁡x  and by squaring,  36(1−sin2x)=8sin2x−811sin⁡x+22  We solve the quadratic equation in sin⁡x and get the two solutions  sin⁡x=411±242326  cos⁡x=22−22sin⁡x6  Anyway, you cant

Mollie Nash · Accepted Answer

Yes it is possible. Write the equation as follows:  (3+i2)e−ix+(3−i2)eix=22  x=2nπ−ilog⁡(±(1±i)1122+3i),   n∈Z  Also, you can use this well-known formula:  arctan⁡x=12ilog⁡(1−ix)−12ilog⁡(1+ix)

Vasquez · Accepted Answer

It is possible to have approximations of the solution. Consider that you look for the zero&#039;s of function f(x)=6cos⁡x+22sin⁡x−22 Since we know at least the exact trigonometric values of multiples of π24, in the considered range, it is easy to show that 9π24&amp;lt;x1&amp;lt;5π12 and 45π24&amp;lt;x2&amp;lt;23π12 If you do not want to use a purely numerical method such as Newton which would work like a charm; consider the infinite series representation f(x)=6cos⁡(a)+22sin⁡(a)−22+∑n=1∞6cos⁡(a+nπ2)+22sin⁡(a+nπ2)n!(x−a)n Truncate the expansion to any order and use series reversion. For example, use the terms up to (x−a)6 For a=9π24 you will obtain an explicit reult (too long to be written here) and its numerical evaluation gives x1=1.2259088208 while the exact solution is x1=1.2259088264 For a=45π24, you will obtain x2=5.9382978047 while the exact solution is x2=5.9382978068

Solve for x which is in [0,2\pi] 6\cos x+2\sqrt2 \sin x=\sqrt{22} I

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