How Do You Solve the Equation 4 Sin x + 2 Cos x = 3 on the Interval [0, 2 Pi]?

[Nadime]

Hi, i signed up for a math exam in Norway come may, and i have to educate myself, because there are no lessons involved, only an exam, but here goes;

Its basic, you have an equation; 4 sin x + 2 cos x = 3 x range is [0, 2 pi]

I guess this is basic stuff, so don't laugh, i couldn't find any examples of this kinda equation in my textbooks, and when i found it in an exam preview i got kinda stressed. I tried to apply some of the different rules of pytagoran identities and such, but with no success.

A friend of mine used this forum for his ib studies, and said that most ppl here rock at math and were really friendly as well, so please help me out :)

 

[Nadime]

well, never mind, i solved it myself!

 

[tacman]

Hi there! Don't worry, we are here to help you out with your math problem. Let's take a look at the equation you provided: 4 sin x + 2 cos x = 3. First, let's rearrange the equation to make it easier to work with: 4 sin x = 3 - 2 cos x. Now, we can use the Pythagorean identity sin^2 x + cos^2 x = 1 to rewrite the equation as 4 sin x = 3 - 2(1 - sin^2 x). Simplifying this, we get 4 sin x = 5 - 2 sin^2 x. This is starting to look more manageable! Now, we can move all the terms to one side and factor out sin x to get sin x(2 sin x + 4) = 5. From here, we can divide both sides by (2 sin x + 4) to isolate sin x and get sin x = 5/(2 sin x + 4). Using the given range for x, we can now plug in values for x and solve for sin x. Once we have the value for sin x, we can use the inverse sine function to find x. I hope this helps and good luck on your exam!