tan2x*cotx - 3 = 0
We know that: tan2x = sin2x/cos2x and cotx = cosx/sinx
==> sin2x/cos2x *cosx/sinx = 3
Now we know that sin2x = 2sinx*cosx
==> 2sinxcosx/cos2x * cosx/sinx = 3
Reduce sinx:
==> 2cos^2 x/ cos2x = 3
Now we know that cos2x = 2cos^2 x-1
==> 2cos^2 x/(2cos^2 x -1) = 3
==> 2cos^2 x = 3(2cos^2 x -1)
==> 2cos^2 x = 6cos^2 x - 3
==> -4cos^2 x= -3
==> 4cos^2 x = 3
==> cos^2 x = 3/4
==> cosx = +-sqrt3/ 2
<span>==> x = pi/6, 5pi/6, 7pi/6, and 11pi/6</span>
What's the question? what's the question?
Answer:
1:+35
2:-270
3:-12
4+19
5:-15
6:-25
Step-by-step explanation:
Answer:
2.256 x 10^25
Step-by-step explanation:
3.9 · 0.5 + 4 5/6 ÷ 3 3/7
1.95 + 4 5/6 ÷ 3 3/7 multiplication left ot right
1.95 + 29/6 ÷ 24/7 convert to improper fraction
1.95 + 29/6 · 7/24 divide by fraction same as multiplication by reciprocal
1.95 + 203/144 multiply
1.95 + 1.40 (and a teensy bit more)
3.35
