Question

d

Answer 1

its just a quadratic in disguise

Answer 2

\[14\sin X \cos X-14\sin X=-7,2\sin X \cos X=-1+2\sin X\]
Squaring both sides,
\[4\sin ^{2}x \cos ^{2}x=1+4\sin ^{2}x-4\sin x\]
\[4\sin ^{2}x \left( 1-\sin ^{2}x \right)-1-4\sin ^{2}x+4\sin x=0\]
\[-4\sin ^{4}x+4\sin x-1=0\]

Answer 3

let u = sin(x)

7u^2 - 14u + 2 = -5

Answer 4

is it sin2x orsin ^2 x

Answer 5

:)

Answer 6

so the answer is zero?

Answer 7

what is the solution of:

7u^2 - 14u + 2 = -5

Answer 8

\(\bf 7 sin^2(x) - 14 sin(x) + 2 = -5 \implies 7 sin^2(x) - 14 sin(x) + 7 = 0\\
7[sin^2(x) - 2sin(x) + 1] = 0\\
\textit{let sin(x) = u as amistre64 said}\\
7[x^2-2x+1] = 0\)

Answer 9

7u^2 - 14u + 2 = -5

7u^2 - 14u + 7 = 0

u^2 - 2u + 1 = 0

(u-1)^2 = 0

Answer 10

woops, made x instead of u, anyhow hhehe

Answer 11

let sin(x) = chi :)

Answer 12

chi chi chia

Answer 13

kiimiilee then just solve the quadratic for "u" or sin(x)

Answer 14

"all" solutions will require you to period the both angles by multiples of 2pi