# A strange trigonometric equation

Today,in our class, we received a trigonometric equation

$$\sin^{10}{x}+\cos^{10}{x}=\frac{29}{16}\cos^4{2x}$$

and the question was to find the general solution of this equation. My approach was, at First, trying to show that there were no solutions using inequalities, but I failed. So, my last method was, expanding RHS by binomial theorem, and canceling some terms out, which at last gives a quadratic in $\sin{x}\cos{x}$. But this way was too long.

Can anyone suggest or give a simpler method? I firmly believe there's one trick in ques to make it easier, which I cannot solve.

• Checking in Wolfram Alpha, the difference of both sides simplifies to something with a $cos(4x)$ factor which gives the roots. (The other factor happens to never vanish). But the trick is getting to that expression by hand, and that's not jumping out at me. Jul 28 '14 at 17:38
• Setting $u = \cos^2 x$, one obtains an equivalent fourth-order polynomial equation in $u$, for which there is an explicit formula. But this probably isn't "easier". Jul 28 '14 at 17:46
• yes, its more complicated, I suppose that just daring to solve by expanding LHS Jul 28 '14 at 17:47
• Conversion to complex exponentials is usually helpful in cases like this. Here, the exponential form of $16\sin^{10} x + 16\cos^{10}x - 29 \cos^4 2x$ decomposes into straightforward factors.
– Blue
Jul 28 '14 at 17:48
• sorry @Blue sir, I didn't get it properly. Can you please tell a bit more? Jul 28 '14 at 17:51

$\displaystyle\cos2x=1-2\sin^2x=2\cos^2x-1$

Setting $\displaystyle\cos2x=u,$ we get $$\left(\frac{1-u}2\right)^5+\left(\frac{1+u}2\right)^5=\frac{29}{16}u^4$$

$$\iff2\left[1+\binom52u^2+\binom54u^4\right]=2^5\cdot\frac{29}{16}u^4$$

Again, $\displaystyle u^2=\frac{1+\cos4x}2$

$$\sin^{10}x + \cos^{10}x = \frac{29}{16} \cos^4 2x$$

We can do some algebraic manipulation with the LHS in the following manner -

$$\sin^{10}x + \cos^{10}x$$

$$\left(\frac{1- \cos 2x}{2}\right)^5 + \left(\frac{1+ \cos 2x}{2}\right)^5$$

$$2\times \frac{\left( \binom{5}{0} + \binom{5}{2} \cos^{2}2x + \binom{5}{4} \cos^4 2x \right)}{2^5}$$

Equate this to the RHS, and solve for the quadratic in $\cos^2 2x.$

We know that $X^5+Y^5=(X+Y)(X^4-X^3Y+X^2Y^2-XY^3+Y^4)$; if we set $X=\sin^2x$ and $Y=\cos^2x$, the left hand side becomes $$(\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)$$ or $$\sin^8x+\cos^8x+\sin^4x\cos^4x-\sin^2x\cos^2x(\sin^4x+\cos^4x)$$ The first three terms can be written $$(\sin^4x+\cos^4x)^2-\sin^4x\cos^4x$$ and $$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x$$ so we get $$(1-2\sin^2x\cos^2x)^2-\sin^4x\cos^4x-\sin^2x\cos^2x(1-2\sin^2x\cos^2x)$$ which further simplifies into $$1-4\sin^2x\cos^2x+4\sin^4x\cos^4x-\sin^4x\cos^4x -\sin^2x\cos^2x+2\sin^4x\cos^4x$$ that is $$1-5\sin^2x\cos^2x+5\sin^4x\cos^4x$$ or $$1-\frac{5}{4}\sin^2(2x)+\frac{5}{16}\sin^4(2x)$$ Can you go on?

• I had the same method, and as stated earlier in my ques, I too arrived at this quadratic in $\sin{x}\cos{x}$, but I felt that's too long to solve, Jul 28 '14 at 17:49
• @Dinesh You can express $\cos^2(2x)$ in terms of $\sin^2(2x)$, can't you? This becomes a biquadratic in $\sin^2(2x)$. Jul 28 '14 at 17:51
• Let me see if that biquadratic is solvable by easy means. Else, it wont give nice results. Jul 28 '14 at 17:53
• @Dinesh I get $\sin^2(2x)=13/12$ (no solutions) or $\sin^2(2x)=1/2$. But check my computations. Jul 28 '14 at 18:00
• wolframalpha.com/input/… Jul 28 '14 at 18:05

can this approach be somehow helpful? set $sin^2x=u$ and $cos^2x=1-u$ similar to what is suggested in one of the comments above. Then, after some manipulations the equations simplify to $$-\frac{(48a^2 - 48a + 13)\cdot(8a^2 - 8a + 1)}{16}=0$$ (I used matlab's symbolic toolbox to obtain this) whose real solutions are $$u=\frac{\sqrt{2}}{4}+\frac{1}{2}$$ and $$u=\frac{1}{2}-\frac{\sqrt{2}}{4}$$ can you continue from here?

• Well this ques was asked to be done in class,, so I dont think ur method would be much useful, but thanks anyways Jul 28 '14 at 18:27

Rewrite $\cos(2x)=1-2\sin^2(x)$ and $\cos^{10}(x)=(1-\sin^2(x))^5$ and then factor $\sin^{10}(x)+(1-\sin^2(x))^5-\frac{29}{16}(1-2\sin^2(x))^4$ to get $$-\frac{1}{16}(48\sin(x)^4-48\sin(x)^2+13)(8\sin(x)^4-8\sin(x)^2+1)=0.$$ The first equation has no solutions since the discriminant is negative and the second has the solutions $\sin(x)=\pm\frac{1}{2}\sqrt{2-\sqrt{2}}$ giving the solutions $\frac{k\pi}{8}, k=\pm 1, \pm 3$.