Trigonometry / Solve for y

Find the value of y such that $$\tan(y°)=(4\cos^29°-3)(4\cos^227°-3)$$

My work

I let $$x = 9°$$ And equation become $$(2\cos2x-1)(2\cos6x-1)$$

Then I tried to convert cos in tan using half angle formula but it didn't work .

I also found $$\cos(4x+6x)=0 \implies \tan4x\tan6x=1$$ But I failed to convert in a form from where I can calculate y.

Thanks in advance for any help

$$\cos3x=\cos x(4\cos^2x-3)$$

For $$\cos x\ne0,$$

$$4\cos^2x-3=\dfrac{\cos 3x}{\cos x}$$

$$(4\cos^29^\circ-3)(4\cos^227^\circ-3)=\dfrac{\cos27^\circ}{\cos9^\circ}\cdot\dfrac{\cos81^\circ}{\cos27^\circ}$$

Finally $$\cos81^\circ=\sin?^\circ$$

• Sorry sir for my ignorance ! But can you explain how $$cos4x=cosx(4cos^2x-3)$$ – Rishi Nov 11 '19 at 10:06
• – lab bhattacharjee Nov 11 '19 at 10:08
• Thank you I know formula , but got confused when you wrote 4x , thanks – Rishi Nov 11 '19 at 10:11
• @Rishi, Sorry for the horrible typo – lab bhattacharjee Nov 11 '19 at 10:20