How do you find the minimum of the equation sin^2(x) + 4sin(x)?

The first step to solving this problem is to treat it as a normal quadratic equation; if you are struggling with comparing our equation to a normal quadratic, try substituting sin(x) = y into the equation as shown:

sin2(x) + 4sin(x) becomes y+ 4y 

Even though our equation does not equal 0 we can still use the 'complete the square' method to help us find the minimum, after applying this method our equation becomes:

(y + 2)2 - 4

From this we can substitue y for sin(x), giving:

(sin(x) + 2)2 - 4 

To find the minimum of our equation we have to take in to account the fact that sin(x) has a range of -1 to 1, which limits (sin(x) + 2) to a range of 1 to 3. 

From this you should be able to deduct that the smallest value of (sin(x) + 2)- 4 is -3. This occurs when sin(x) = -1. 

(-1 + 2)- 4 = 1- 4 = -3

Hence the minimum of sin2(x) + 4sin(x) is -3. 

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