2cos^2(x) + 3sinx = 0?

Question

find all exact real solutions x

thanks!

Answer 1

replacae cos^2 with 1-sin^2 x

2(1-sin^x) + 3sinx = 0
-2sin^2x + 3sinx + 2 = 0

2sin^2x -3sinx - 2 = 0
(2sinx + 1)(sinx - 2) = 0
sinx = -1/2 or sinx =2 ---but sinx cannot = 2 so just work with -1/2

sinx = -1/2 quadrants III and IV
x = 210 deg (or 7pi/6) + 2npi in QIII
x = 330 deg (or 11pi/6) + 2npi in QIV

Answer 2

Substitute 1-sin^2x = cos^2x getting
2(1-sin^2x) +3 sinx = 0
-2sin^2x +3sinx +2 = 0
sinx = {-3 +/- 5}/-4
sin x = -8/-4 = 2 which is impossible and is rejected
sinx = 2/-4 = -1/2
x = arcsin(-1/2) which gives solutions in quadrants III & IV.
Each solution has additional solutions by adding or subtracting 2npi where n is any integer.

Answer 3

2(1 - sin^2 (x)) + 3sin(x) = 0
2sin^2 (x) -3sin(x) -2 = 0
(2sinx+1)(sinx-2) = 0
Thus, sinx = -1/2 because sinx = 2 has no solutions; as a result, x = 210 or 330 degrees (these are the only solutions between 0 and 360).

Answer 4

1(1-sin^2(x))+3sin(x)=0.... Replace cos^2(x) with (1-sin^2(x))

-sin^2(x)+3sin(x)+1=0 Distribute and rearrange the terms


Let w=sin(x)


-w^2+3w+1=0



Now use the quadratic equation to solve for w. Remember, you want to solve for x. So just substitute the answer in terms of w into w=sin(x)

Answer 5

My sister can do it