Question

Solve each equation giving a general formula for all of the solutions.
1. 2sin^2 x - 5sinx - 3 = 0
2. (cosx - √(2)/2)(secx - 1) = 0
I think the answer for #1 is x = 7π/6 + 2πn, x = 11π/6 + 2πn
#2 are x = 7π/4 + 2πn

Answer 1

How did you factor it?

Answer 2

Only one answer for 1.

Answer 3

two answers for 2.

Answer 4

You can substitute sin(x) for a variable and factor the resulting equation right?

Answer 5

\[2\sin ^{2}x-5\sin x-3=0\]
\[2u ^{2}-5u-3=0\]
Like this, and then solve it?

Answer 6

yes

Answer 7

The 2nd one is already factored so solve each factor for zero using inverse functions.

Answer 8

I still got x = 7π/6 + 2πn and x = 11π/6 + 2πn for #1

Answer 9

Right, there's one factor two answers involving the sine function. Did you get the second one?

Answer 10

one useful factor for the first but u got it.

Answer 11

I got x = 7π/4 + 2πn and x = π/4 + 2πn