Hi I need some help with this equation.

cos(2 theta + 0.2) = -0.2
I start by taking arccos of both sides
cos^-1(-0.2) = 2 theta + 0.2
cos^-1(-0.2)-0.2 = 2 theta
(cos^-1(-0.2)-0.2 )/ 2 = theta

I got theta as 0.786

then to work out the additional values

2pi - 0.786 = 5.49
5.49 - 2pi = -0.786

0.786 is a correct solution, however another solution is -0.986
Can someone explain how I get to -0.986

Hi I need some help with this equation cos2 theta 02 02 I start by taking arccos of both sides cos102 2 theta 02 cos10202 2 theta cos10202 2 theta I got theta a class=

Respuesta :

!<Answer>!

To understand how to get the additional solution of -0.986 for theta, let's review the steps you followed:

1. You started by taking the inverse cosine (arccos) of both sides of the equation: cos^(-1)(-0.2) = 2theta + 0.2.

2. You then subtracted 0.2 from both sides of the equation: cos^(-1)(-0.2) - 0.2 = 2theta.

3. Next, you divided both sides of the equation by 2 to isolate theta: (cos^(-1)(-0.2) - 0.2) / 2 = theta.

Based on these steps, you correctly found the value of theta as 0.786.

To find the additional solution, we need to consider the periodic nature of the cosine function. The cosine function has a period of 2π, meaning it repeats every 2π radians.

Since you obtained 0.786 as the initial solution for theta, you can find the additional solution by subtracting the period, 2π, from the initial solution.

0.786 - 2π = -0.986.

Therefore, -0.986 is the additional solution for theta. By subtracting 2π from the initial solution, you accounted for the periodicity of the cosine function and found the second possible value for theta.

~ Sun

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