Question

Solve these two equations :
2sin^2 x - 5sin x - 3 = 0
(cos x - (square root of 2)/2)(sec x - 1) = 0

Answer 1

for the first one, let u = sin(x)

you now have to solve the quadratic:

2u^2 - 5u - 3 = 0

when you solve for u, replace it back with sinx

Answer 2

2*-3=-6
6-1=5
6*-1=-6
\[2 \sin ^2x-\left( 6-1 \right)\sin x-3=0,~or~2~\sin ^2x-6\sin x+\sin x-3=0\]
make factors and solve

Answer 3

as Euler271 and surjithayer suggested \(\bf 2{\color{blue}{ sin}}^2{\color{blue}{ (x)}} - 5{\color{blue}{ sin(x)}} - 3 = 0\qquad {\color{blue}{ u=sin(x)}}
\\ \quad \\
2{\color{blue}{ u}}^2-5{\color{blue}{ u}}-3=0\)

Answer 4

so factor like 2x^2 - 5x - 3?

Answer 5

\[2 \sin x \left( \sin x-3 \right)+1\left( \sin x-3 \right)=0\]
\[\left( \sin x-3 \right)\left( 2 \sin x+1 \right)=0\]
either sin x-3=0, sin x=3,Rejected,because
\[\left| \sin x \right|\le 1\]
\[\sin x=-\frac{ 1 }{2 }=-\sin 60=\sin \left( 180+60 \right),\sin \left( 360-60 \right)\]
calculate x

Answer 6

yes, as surjithayer said already, keep in mind that \(\bf -1\le sin(x) \le 1\)

Answer 7

ok now I've got that I've got to factor 2x^2 - 5x - 3

Answer 8

(x - 3) (2x+1)

Answer 9

@surjithayer

Answer 10

x would represent sin^2x? or just sin x? in the factorization

Answer 11

anyone there?

Answer 12

@sourwing

Answer 13

x=sin x