Question

Find all solutions to the equation.

7 sin2x - 14 sin x + 2 = -5

Answer 1

rewrite the equation and its

\[7\sin^2(x) -14x +7 = 0\]

divide each term by 7

\[\sin^2(x) -2\sin(x) + 1 = 0\]

factor the perfect square

then solve each factor for the solution.

Answer 2

ok

Answer 3

it some times is easy to think of sin(x) is some variable a

then
7a^2-14a+2=-5

Answer 4

But what are all the possible solutions ?

Answer 5

7a^2-14a+7=0
7a^2-7a-7a+7=0
7a(a-1)-7(a-1)=0
(7a-7)(a-1)=0
so
a=1
s0 sin(x) = 1
when does
sin(x) = 1?

Answer 6

pi/2 +- 2pi*k where k runs through the integers.

Answer 7

I am confused should that be \[7\sin^2x -14sinx +2=5\]?

Answer 8

yes

Answer 9

Yeah thats it

Answer 10

If so it would be easiest to let \[u = sinx\] then the equation is\[7u^2 - 14u +2 = -5\]. Now add 5 to both sides of the equation and factor out the 7. This results in
\[7u^2-14u+2=5\]
\[7u^2-14u+7=0\]
\[7(u^2-2u+1)=0\]
\[u^2-2u+1=0\]
This is the common quadratic trinomial that can be factored to \[(u-1)^2=0\]
So we need to determine when u = 1.
Since \[u = \sin x\], we are looking for any time where sin x = 1.
This is when \[x = \frac{ \pi }{ 2 }+2\pi n\] where \[n \in \mathbb{Z}\]

Answer 11

Substituting a variable in for a trig function to make it easier to solve and then re-substituting the trig function later after solving the equation is a common method of solving trigonometric equations. It is also used quite often in calculus.

Answer 12

I hope that helps.

Answer 13

Yeah thanks