Question

Find all solutions to the equation.

7sin^2 x - 14sin x + 2 = -5

Answer 1

Move the -5 to the left.
Then you have a quadratic in sin x

Answer 2

So you get 7sin^2 x -14sin x + 7 = 0 and that turns into

x(7sin - 7)(sin - 1)?

Answer 3

No.
Treat sin x as if it were a variable.
One method to avoid confusion is to do a substitution.

Let \(u = \sin x\).
Then \(u^2 = \sin^2 x\).

Now rewrite your quadratic equation substituting the u's above for the sines.

Answer 4

7u^2x - 14ux + 7 = 0

could you divide by 7 here?

Answer 5

\(7\color{red}{\sin^2 x} - 14\color{green}{\sin x} + 7 = 0\)

Substitutions:

\(\color{red}{u^2 \sin^2 x }\)

\(\color{green}{u = \sin x}\)

You now have:

\(7u^2 - 14u + 7 = 0\)

and you have a simple quadratic equation in u.
Now you solve. You can try factoring or the quadratic formula to solve for u.

Answer 6

You are close, but notice that u = sin x, so when you substitute, there are no x's left.
The u's take over completely as the variable.

Answer 7

You are correct about dividing by 7.
Take out the common factor of 7.

Answer 8

(7u - 7)(u - 1) so u=1

you also get this if you divide by 7: (u - 1)(u - 1) u=1 in each

Answer 9

Above in red it should read:

\(\color{red}{u^2 = \sin^2 x}\)

I accidentally left out the equal sign.

Answer 10

Correct.
The substitution allowed you to solve the quadratic and get the single solution u = 1 of the quadratic equation.
Good work so far.

Now we have to undo the substitution.
Since we originally let u = sin x, now we substitute the sine of x back to get:

\(u = 1\)

\(\sin x = 1\)

Answer 11

Now we need to solve the trig part of the equation.

For which values of x is \(\sin x = 1\)?

Answer 12

x= pi/2, 5pi/2, 9pi/2, -3pi/2, -7pi/2 and so on. It says list them all and there are unlimited.