Question

Step 2 Treat the resulting equation as a quadratic in sec(?) and factor the left side of the equation. -2 cos^2(?) - cos(?) + 1 = 0 2 cos^2(?) + cos(?) - 1 = 0 (2 cos(?) - 1)(2 sec(?) - 1 x cos(?) + 1) = 0 Step 3 Set each factor equal to zero and solve for cos(?). ? 2 cos(?) - 1 = 0 2 cos(?) = cos(?) = / 2 ? cos(?) + 1 = 0 cos(?) =

          Step 2
Treat the resulting equation as a quadratic in sec(?) and factor the left side of the equation.
-2 cos^2(?) - cos(?) + 1 = 0
2 cos^2(?) + cos(?) - 1 = 0
(2 cos(?) - 1)(2 sec(?) - 1 x cos(?) + 1) = 0
Step 3
Set each factor equal to zero and solve for cos(?).
? 2 cos(?) - 1 = 0
2 cos(?) =
cos(?) = / 2
? cos(?) + 1 = 0
cos(?) =

Step 2
Treat the resulting equation as a quadratic in sec(?) and factor the left side of the equation.
-2 cos^2(?) - cos(?) + 1 = 0
2 cos^2(?) + cos(?) - 1 = 0
(2 cos(?) - 1)(2 sec(?) - 1 x cos(?) + 1) = 0
Step 3
Set each factor equal to zero and solve for cos(?).
? 2 cos(?) - 1 = 0
2 cos(?) =
cos(?) = / 2
? cos(?) + 1 = 0
cos(?) =

Added by Gregory W.

Introductory and Intermediate Algebra for College Students 4th

Robert Blitzer 4th Edition

Instant Answer

Solved by Expert Julie Silva

04/18/2020

Step 1

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Treat the resulting equation as a quadratic in sec(θ) and factor the left side of the equation. -2 cos^2(θ) - cos(θ) + 1 = 0 2 cos^2(θ) + cos(θ) - 1 = 0 (2 cos(θ) - 1)(2 sec(θ) - 1 cos(θ) + 1) = 0 Set each factor equal to zero and solve for cos(θ). 2 cos(θ) - 1 = 0 2 cos(θ) = cos(θ) = / 2 cos(θ) + 1 = 0 cos(θ) =