Cos(3x + (π)/(6)) = -1 2cos^2 3β + sin 3β = 1

Name: cos(3x + (π)/(6)) = -1 2cos^2 3β + sin 3β = 1
Uploaded: 2021-01-12T19:59:07-08:00
Duration: 4 min 30 s
Channel: Pawan Yadav
Description: cos(3x + (π)/(6)) = -1 2cos^2 3β + sin 3β = 1

Transcript

00:01 Okay, so we are said to find all the angles in degrees satisfying the given equations.

00:06 Okay, so let's start with first one that is cos 3x is equal to 1.

00:11 And we know cost is 1 when we have angle as 0.

00:15 Okay, so let's write for cost to become 1.

00:20 So that is possible when theta is equals to 0.

00:24 Then at pi it's minus 1 then it's 2 pi and it's 4 pi so in general we can write this as 2 n pi right and here pi is the angle in radian which is equals to 180 degree so this theta is 2 n pi so therefore 3x can be written as on comparing this we get 2 n pi and x is 2 by 3 n pi sorry n where pi is if we write in degrees so this will be n times 180 degrees okay and here we have to remember n can be any integer value okay so that is a solution for x now second sine 2x equals to 0 so again we know sine theta to be 0 theta must be 0 sign 0 is 0 then at pi it is 0 then at pi it is 0 then at 2 pi and it continues.

01:28 So this can be written in general theta is n pi, right, where n belongs to any integer.

01:36 Okay.

01:37 So now we are given that theta is 2x, so 2x is n pi and therefore x must be n pi by 2, but we have to write the angle in degrees.

01:47 So we know pi is equal to or pi corresponds to 180 degrees.

01:54 So that is our answer with n belongings to integer.

01:58 Okay, so we are done with 2.

02:00 Now third, sine is root 2 by 2...

Question

\(\cos\left(3x + \frac{\pi}{6}\right) = -1\) \(2\cos^2 3\beta + \sin 3\beta = 1\)

Question

\(\cos\left(3x + \frac{\pi}{6}\right) = -1\) \(2\cos^2 3\beta + \sin 3\beta = 1\)

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