Question

If sin2x=22 - 8√7 Prove that 3(Sinx+Cosx+Tanx+Cosecx+Secx+Cotx) =21
If you get correct, You will get 1 owl buck, fan, and medal!

Answer 1

@misty_fox
@AravindG
@inkyvoyd
@Electric_Blue
@Mehek14
@Rushwr
@AloneS
@Serenity74
@tootzrll
@boldjon
@perl
@mathmale

Answer 2

too lazy

Answer 3

lol, this is actually a solveable equation. and the answer is simple

Answer 4

I know but I'm lazy

Answer 5

@Anaise
@ChantySquirrel1129**
@CandyCove
@imqwerty
@Zeronknight
@Somy

Answer 6

@myah

Answer 7

Really i was last LOL it's cool ill keep working on it

Answer 8

@Safa102

Answer 9

Let sin 2x = k

k = 22- 8√7

k - 22 = -8√7 [Bring 22 to the LHS)

(k-22)² = (-8√7)² [Squaring both sides]

k² - 44k + 484 = 448 [Expanding]

k² - 44k = 448-484 [Bringing 484 to the RHS]

k² - 44k = -36 [Multiply throughout by k]

k³ - 44k² = -36k

k³ - 44k² + 36k = 0 [Bringing -36k to the LHS]

k³ - 49k² +5k² + 8k + 28k = 0 [Rearranging factors]

k³ + 5k² + 8k = 49k² - 28k [Rearranging factors and bring 49k² - 28k to the RHS]

k³ + 4k² + k² + 4k + 4k = 49k² - 28k [Rearranging factors]

k³ + 4k² + k² + 4k + 4k + 4 - 4 = 49k² - 28k [Adding + 4 and -4 to make it a quadratic/polynomial]

k³ + 4k² + k² + 4k + 4k + 4 = 49k² - 28k + 4 [Bringing -4 to the RHS]

k³ + 4k² + k² + 4k + 4k + 4 = (7k - 2)² [Writing 49k² - 28k + 4 in the form of (a-b)²]

(1+k)(k² + 4k + 4) = (7k - 2)² [Factoring LHS]

(1+k)(k + 2)² = (7k - 2)² [Writing k² + 4k + 4 in the form of (a+b)²]

(1+k) = (7k-2)²/(k+2)² [Bringing (k+2)² to the RHS]

(1 + 2sinxcosx) = (7k-2)²/(k+2)² [k = sin2x = 2sinxcosx]

(sin²x + cos²x + 2sinxcosx) = (7k-2)²/(k+2)² [ 1 = sin²x + cos²x and we bring LHS to the form (a+b)²]

(sinx + cosx)² = (7k-2)²/(k+2)²

(sinx + cosx) = (7k-2)/(k+2) [Taking root on both sides]

sinx + cosx = (7k -2)/k / (k+2)/k) [Divide Numerator and Denominator of RHS by k]

sinx + cosx = (7 - 2/k) / (1+2/k)

(sinx + cosx) (1 + 2/k) = 7 - 2/k [Taking (1+2/k) to the LHS]

sinx + cosx + 2sinx/k + 2cosx/k = 7 - 2/k [Multiplying and Expanding]

sinx + cosx + 2/k(sinx + cosx) + 2/k = 7 [Taking Common factor out]

sinx + cosx + 2/k(sinx + cosx + 1) = 7 [Taking Common factor out]

sinx + cosx + 2/k(sinx + cosx + sin²x + cos²x) = 7 [ 1 = sin²x + cos²x]

sinx + cosx + 2/2sinxcos (sinx + cosx + sin²x + cos²x) = 7 [k = sin2x = 2sinxcosx]

sinx + cosx + 1/cosx + 1/sinx + sinx/cosx + cosx/sinx = 7 [Canceling common terms]

sinx+cosx+secx+cosecx+tanx+cotx = 7 [1/cosx=secx,1/sinx=cosecx,sinx/cosx=tan...

Therefore sinx + cosx + secx + cosecx + tanx + cotx = 7

3(sinx + cosx + secx + cosecx + tanx + cotx) = 21
MY SQRT SYMBOLS DISAPEARED!