Question

How many values of $x$ in the interval $[0,2\pi ]$ satisfies the equation ${\sin }^{6}x=1+{\cos }^{4}3x$

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Answer 1

Maximum value of ${\sin }^6$ x is 1 and minimum value of 1+ ${\cos }^4$ 3x is 1.

$\Rightarrow$ ${\sin }^6$ x $\le$1 and ${\sin }^6$x = 1 + ${\cos }^4$ 3x $\ge$ 1

$\Rightarrow$ 1$\le$ ${\sin }^6$ x $\le$ 1

$\Rightarrow$ ${\sin }^6$ x = 1

$\Rightarrow$ ${\cos }^4$ 3x = ${\sin }^6$ x - 1 = 1 - 1 = 0

$\Rightarrow$ ${\sin }^6$ x = 1 and ${\cos }^4$ 3x = 0

$\Rightarrow x=(2n+1)\frac{\pi }{2}\text{and}3x=(2m+1)\frac{\pi }{2}.$
$\Rightarrow x=(2n+1)\frac{\pi }{2}\text{and}x=(2m+1)\frac{\pi }{6}.$
We want to find the common values of $(2n+1)\frac{\pi }{2}and(2m+1)\frac{\pi }{6}$. For some value of m and n,

both these expressions will be equal corresponding to common terms.
$\Rightarrow (2n+1)\frac{\pi }{2}=(2m+1)\frac{\pi }{6}$

$\Rightarrow 2m+1=6n+3$

$\Rightarrow 2m=6n+2$
$\Rightarrow$ m = 3n + 1

This means for any value of n, there will be a corresponding m.

$\Rightarrow$ All the values of (2n + 1) $\frac{\pi }{2}$ is also part of the sequence (2m + 1) $\frac{\pi }{6}$.

So the solution is x = (2n + 1) $\frac{\pi }{2}.$

The values in the interval [0, 2π] are $\frac{\pi }{2}and\frac{3\pi }{2}$ corresponding to n = 0 and n = 1.

So the number of solutions in the interval [0, 2π] is 2.