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Today,in our class, we received a trigonometric equation


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.

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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. – Semiclassical 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". – A Blumenthal Jul 28 '14 at 17:46
yes, its more complicated, I suppose that just daring to solve by expanding LHS – Dinesh 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? – Dinesh Jul 28 '14 at 17:51


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


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

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

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 $\cos^2 2x.$

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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?

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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, – Dinesh 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 $\sin^2(2x)$. – egreg Jul 28 '14 at 17:51
Let me see if that biquadratic is solvable by easy means. Else, it wont give nice results. – Dinesh 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. – egreg Jul 28 '14 at 18:00… – Dinesh 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?

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Well this ques was asked to be done in class,, so I dont think ur method would be much useful, but thanks anyways – Dinesh 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 $\frac{k\pi}{8}, k=\pm 1, \pm 3$.

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