How do you solve $2cos^2(x/3) + 3sin(x/3) - 3 =0$ ?

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Answer 1 · 2015-05-06T16:09:12

Cal $sin (x/3) = t.$
$y = 2(1 - t^2) + 3t - 3 = 0$
Quadratic equation.

$-2t^2 + 3t - 1 = 0$

Since a + b + c = 0, one real root is t= 1 and the other is (t = 1/2)

a. $t = sin (x/3) = 1 -> x/3 = pi/2 -> x = (3pi)/2$

b. $sin (x/3) = t = 1/2 -> x/3 = pi/6 -> x = (3pi)/6 = pi/2$

c. $sin (x/3) = 1/2 -> x/3 = (5pi)/6 -> x = (15pi)/6 = (pi/2 + 2pi).$

Within interval (0, 2pi), there are 2 answers: $pi/2 and (3pi)/2.$

Always check the answers.

$x = pi/2 -> x/3 = pi/6 -> y = 2.(3/4) + 3.(1/2) - 3 = 3/2 + 3/2 - 3 = 0$ . Correct
$x = (3pi)/2 -> x/3 = pi/2 -> y = 0 + 3 - 3 = 0$ Correct.

How do you solve 2cos^2(x/3) + 3sin(x/3) - 3 =02cos^2(x/3) + 3sin(x/3) - 3 =0?