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4cos(x)+ 3 = 6 with - π< x < π​

User Schlamar
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1 Answer

3 votes

Answer:

x = arccos(3/4) + 2πn where n = 0,1,2,3,4, ....

Explanation:

We can solve the equation 4cos(x) + 3 = 6 by isolating the cos(x) term on one side of the equation.

4cos(x) + 3 - 3 = 6 - 3

4cos(x) = 3

cos(x) = 3/4

Now we can find the solutions of the equation by finding the values of x that make cos(x) equal to 3/4. Since -π < x < π, we can find the solutions in the range of -π < x < π.

x = cos^-1(3/4) + 2πn, where n is an integer

x = cos^-1(3/4) + 2πn, where n = 0, 1, 2, ...

The solutions are x= cos^-1(3/4) and x = cos^-1(3/4) + 2π, x = cos^-1(3/4) + 4π

Keep in mind that cos^-1(3/4) is the inverse cosine function or arccos(3/4) so the final solution is x = arccos(3/4) + 2πn where n = 0,1,2,3,4, ....

User Glenda
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