SOLUTION: What are the solutions from 0 degrees to 360 degrees for problem 3a and 5a? 1a. 2 sin^2 x - 1 = 0 Answer: x = 45°, 135°, 225°, 315° ? 3a. 2 sin^2

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: What are the solutions from 0 degrees to 360 degrees for problem 3a and 5a? 1a. 2 sin^2 x - 1 = 0 Answer: x = 45°, 135°, 225°, 315° ? 3a. 2 sin^2       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1140081: What are the solutions from 0 degrees to 360 degrees for problem 3a and 5a?
1a. 2 sin^2 x - 1 = 0
Answer: x = 45°, 135°, 225°, 315° ?


3a. 2 sin^2 x + sin x = 1


5a. 2 sin^2 x + 7 sin x = 4
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

3a.
2sin%5E2+%28x%29+%2B+sin+%28x%29+=+1+
2%28sin%5E2+%28x%29+%2B+%281%2F2%29sin+%28x%29+%29=+1.......complete square
2%28sin%5E2+%28x%29+%2B+%281%2F2%29sin+%28x%29+%2Bb%5E2%29-2b%5E2=+1
2%28sin%28x%29+%2B+%281%2F4%29%29%5E2-2%281%2F4%29%5E2=+1
2%28sin%28x%29+%2B+%281%2F4%29%29%5E2-2%281%2F16%29=+1
2%28sin%28x%29+%2B+%281%2F4%29%29%5E2-%281%2F8%29=+1
2%28sin%28x%29+%2B+%281%2F4%29%29%5E2=+1%2B1%2F8
2%28sin%28x%29+%2B+%281%2F4%29%29%5E2=+9%2F8
%28sin%28x%29+%2B+%281%2F4%29%29%5E2=+9%2F16
sin%28x%29+%2B+%281%2F4%29=+sqrt%289%2F16%29
sin%28x%29+%2B+%281%2F4%29=+3%2F4
sin%28x%29+=+3%2F4-+1%2F4
sin%28x%29+=+1%2F2
x=pi+%2F6%2B2pi+%2An, x=5pi%2F6%2B2pi+%2An
or
x=30°
x=150°

5a.
2sin%5E2+%28x%29+%2B+7+sin%28+x%29+=+4 .......let sin%28x%29=u
2u%5E2%2B+7+u-4=0.........use quadratic formula
u=%28-7%2B-sqrt%287%5E2-4%2A2%2A%28-4%29%29%29%2F%282%2A2%29
u=%28-7%2B-sqrt%2849%2B32%29%29%2F4
u=%28-7%2B-sqrt%2881%29%29%2F4
u=%28-7%2B-9%29%2F4
solutions:
u=%28-7%2B9%29%2F4
u=2%2F4
u=1%2F2
u=%28-7-9%29%2F4
u=-16%2F4
u=-4
=> sin%28x%29=1%2F2+
or sin%28x%29=-4

x=sin%5E-1%281%2F2%29 =>x=+30° ,x=150°
x=sin%5E-1%28-4%29 =>+none

Answer by ikleyn(52915) About Me  (Show Source):
You can put this solution on YOUR website!
.

2*sin^2(x) + sin(x) = 1


2*sin^2(x) + sin(x) - 1 = 0


Introduce new variable  u = sin(x).  Then your equation takes the form


2u^2 + u - 1 = 0.


Find the roots using the quadratic formula


u%5B1%2C2%5D = %28-1+%2B-+sqrt%28%28-1%29%5E2+-+4%2A2%28-1%29%29%29%2F%282%2A2%29 = %28-1+%2B-+sqrt%281+%2B8%29%29%2F4 = %28-1+%2B-+sqrt%289%29%29%2F4 = %28-1+%2B-+3%29%2F4%29.


There are two roots:  1)  u = %28-1+%2B+3%29%2F4%29 = 2%2F4 = 1%2F2,  and


                      2)  u = %28-1+-+3%29%2F4%29 = -4%2F4 = -1.


Case 1).   u = 1%2F2  ====>  sin(x) = 1%2F2  ====>  there are 2 solutions for x:  x= 30°  and  x= 150°.



Case 2).  u = -1  ====>  sin(x) = -1  ====>  x= 270°.


ANSWER.  This equation has 3 solutions:  30°,  150°  and  270°.

Solved.

Introducing new variable is the standard method solving such equations.


/\/\/\/\/\/\/\/

Tutor @MathLover1 lost the solution 270°.

Her solution is very good illustration and demonstration what may happen when you don't follow the standard way: 

    - you do tons of unnecessary calculations;

    - then you lose your horizon;

    - then you make mistakes.