MHB Solve the trigonometric equation

[anemone]

Solve the equation $\sin^6 a+\cos^6 a=0.25$.

 

[kaliprasad]

Solve the equation $\sin^6 a+\cos^6 a=0.25$.

Spoiler

using $x^3+y^3 = (x+y)^3 - 3xy(x+y)$ and $a= \sin^2 a$ and $y = \cos^2 a$

we get $\sin^6 a+\cos^6 a = 1- 3\sin^2a \cos^2 a = .25$
or $3\sin^2a \cos^2 a= .75$
or $4\sin^2a \cos^2 a= 1$
or $\sin^22 a = 1$
or $\sin 2a = \pm1$
or $2a = n\pi\pm \frac{\pi}{2}$

hence $a= \frac{ n\pi\pm \frac{\pi}{2}}{2}$

or $a = (\frac{n}{2}\pm\frac{1}{4})\pi$

because we are taking $\pm{1/4}$ from integer and half integer we can simplify above as
$a = (n\pm\frac{1}{4})\pi$

 

[soroban]

Hello, anemone!

Solve: $\;\sin^6\!x+\cos^6\!x\:=\:\frac{1}{4}$

Spoiler

$\text{Factor: }$
$\quad\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}}(\sin^4\!x - \sin^2\!x\cos^2\!x + \cos^4\!x) \:=\:\frac{1}{4}$

$\begin{array}{c}\sin^4\!x- \sin^2\!x(1-\sin^2\!x) + (1-\sin^2\!x)^2 \:=\:\frac{1}{4} \\ \\
\sin^4\!x - \sin^2\!x +\sin^4\!x + 1 - 2\sin^2\!x + \sin^4\!x \:=\:\frac{1}{4} \\ \\3\sin^4\!x - 3\sin^2\!x + \frac{3}{4} \;=\;0 \\ \\
\frac{3}{4}(4\sin^4\!x - 4\sin^2\!x + 1) \;=\;0 \\ \\
(2\sin^2\!x - 1)^2 \;=\;0 \\ \\
2\sin^2\!x - 1 \;=\;0 \\ \\
2\sin^2\!x \:=\:1 \\ \\
\sin^2\!x \:=\:\frac{1}{2} \\ \\
\sin x \:=\:\pm\frac{1}{\sqrt{2}} \\ \\
x \;=\;\begin{Bmatrix}\frac{\pi}{4} + \pi n \\ \frac{3\pi}{4} + \pi n \end{Bmatrix}
\end{array}$

 

[DreamWeaver]

Solve the equation $\sin^6 a+\cos^6 a=0.25$.

Great problem, Anemone! (Sun)I'm sure at least one of the other proofs is the same, but I'm not going to check until after I've posted... (Bandit)

Spoiler

$$\cos^2 a= 1-\sin^2 a$$

Let $$x=\sin^2a, \quad \Rightarrow$$

$$x^3+(1-x)^3 = 1/4 = $$

$$x^3 + (1-3x+3x^2-x^3) = 3x^2-3x+1$$Hence $$3x^2-3x+\frac{3}{4} = 0 \Leftrightarrow$$

$$x= \frac{1}{2}$$

and

$$a = \pm \sin^{-1}\sqrt{x} = \pm \sin^{-1}\frac{1}{\sqrt{2}} =\pm \frac{\pi}{4}$$