Cos² 2x = (1+cos 4x )/2
cos² 6x = (1+cos 12x)/2
subtract to get 1/2 (cos 4x - cos 12x )
= 1/2 (2 sin 4x sin 8x ) as cos A - cos B = 2 [ sin (A+B)/2 sin (A-B)/2 ]
so the answer

4 0

3 years ago

Exercise 6

Lady_Fox [76]

$a)\\\\x^2-2x-4=0\\\\x^2-2x=4\\\\x^2-2x+1^2=4+1^2\\\\(x-1)^2=5\\\\x-1=\pm\sqrt5\\\\\underline{x=\pm\sqrt5+1\approx-1.24\ or\ 3.24}\\\\\\b)\\\\x^2-4x-2=0\\\\x^2-4x=2\\\\x^2-4x+2^2=2+2^2\\\\(x-2)^2=6\\\\x-2=\pm\sqrt6\\\\\underline{x=\pm\sqrt6+2\approx-0.45\ or\ 4.45}$

$c)\\\\x^2+8x+5=0\\\\x^2+8x=-5\\\\x^2+8x+4^2=-5+4^2\\\\(x+4)^2=11\\\\ x+4=\pm\sqrt{11}\\\\\underline{x=\pm\sqrt{11}-4\approx-7.32\ or\ -0.68}\\\\\\x)\\\\x^2+22x-15=0\\\\x^2+22x=15\\\\x^2+22x+11^2=15+11^2\\\\(x+11)^2=136\\\\x+11=\pm\sqrt{136}\\\\\underline{x=\pm\sqrt{136}-11\approx-22.66\ or\ 0.66}$

$e)\\\\x^2-7x+10=0\\\\x^2-7x=-10\\\\x^2-7x+(\frac72)^2=-10+(\frac72)^2\\\\x^2-7x+(\frac72)^2=-10+(\frac72)^2\\\\(x-\frac72)^2=-\frac{20}4+\frac{49}4\\\\ (x-\frac72)^2=\frac{29}4\\\\x-\frac72=\pm\sqrt{\frac{29}4}\\\\ \underline{x=\pm\frac{\sqrt{29}}2+\frac72\approx6.19\ or\ 0.81}\\\\\\f)\\\\x^2+5x-2=0\\\\x^2+5x=2\\\\x^2+5x+(\frac52)^2=2+(\frac52)^2\\\\(x+\frac52)^2=2+6.25 \\\\x+\frac52=\pm\sqrt{8.25}\\\\\underline{x=\pm\sqrt{8.25}-\frac52\approx-5.37\ or\ 0.37}$

$g)\\\\x^2+9x-4=0\\\\x^2+9x=4\\\\x^2+9x+(\frac92)^2=4+(\frac92)^2\\\\(x+\frac92)^2=4+20.25\\\\x+\frac92=\pm\sqrt{24.25}\\\\\underline{x=\pm\sqrt{24.25}-4.5\approx-9.42\ or\ 0.52}$

$h)\\\\x^2-3x-6=0\\\\x^2-3x=6\\\\x^2-3x+(\frac32)^2=6+(\frac32)^2\\\\(x-\frac32)&^2=6+2.25\\\\x-\frac32=\pm\sqrt{8.25}\\\\ \underline{x=\pm\sqrt{8.25}+1.5\approx4.37\ or-1.37}$

3 0

3 years ago

Please can I have some help with these?

LenaWriter [7]

I hope this helps you

4 0

3 years ago

Read 2 more answers