# Explain how solve this hard equation sin^3 x+sin^2 3x+sin x=3?

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### 2 Answers

You need to remember that the sine function take on y values in interval [-1,1], hence, the following inequality holds for sine function, such that:

`-1 =< sin x <= 1 => -1 =< sin 3x <= 1`

Raising to square `sin 3x <= 1` yields:

`sin^2 3x <= 1`

Raising to cube `sin x <= 1` yields:

`sin^3 x <= 1`

Hence, evaluating the given relation yields:

`sin^3 x + sin^2 3x + sin x <= 1 + 1 + 1 = 3 => sin x = 1 => x = sin^(-1) 1 + n*pi => x = (-1)^k*pi/2 + n*pi` **Hence, evaluating the solution to the given equation yields **`x = (-1)^k*pi/2 + n*pi.`

yields:

Hence, evaluating the solution to the given equation yields

IS'NT ENOUGH!

HOW DID YOU GET HERE? MAY YOU EXPLAIN?