Question

Solve the interval [0, 2pi)
2sinTheta^2-cosTheta-1=0

Answer 1

ok

Answer 2

Can you guide me in the direction of solving this equation?
\[2\sin \theta-\cos \theta-1=0\]

Answer 3

Is it:\[2\sin^{2}\Theta - \cos \Theta - 1 = 0\]

Answer 4

hold on

Answer 5

Consider the equation:
cos (2 theta ) = sin (theta)

Answer 6

yes it is \[2\sin \theta ^{2}-\cos \theta-1=0\]

Answer 7

Using identity cos (2 theta ) = 1 - 2 ( sin (theta) )^2, the above equation becomes:
1 - 2 ( sin (theta) )^2 = sin(theta)
which can be rewritten as
2 ( sin (theta) )^2 + sin(theta) -1 = 0

Now this is is a quadratic equation in one variable .Either solve it using the formula for the solutions to a quadratic equation OR use the following method: Above equation can be rewritten as

( sin(theta) + 1 ) ( 2sin(theta) -1 ) = 0
Thus we have
sin(theta)= -1 OR sin(theta)= 0.5
Hence,
theta=pi/6, 5pi/6 OR 3pi/2
That is all.

Answer 8

you get it.

Answer 9

Thank you, I was wondering If I could solve it using my identities with or without switching to the RHS. Thank you for clearing it up for me! :)

Answer 10

no prob