2cos^2 theta + 5 sin theta - 4 = 0
can give me a further and detail explanation where did the theta on the 5 and the formula on changing the cos theta.
you know that cos^2 = 1-sin^2
the rest is just algebra I
I dropped the thetas just for ease of typing.
Naturally, sin can never be 2, so that's a non-solution, even though it does satisfy the original equation.
To solve the equation 2cos^2(theta) + 5sin(theta) - 4 = 0, we can use the trigonometric identities and algebraic methods. Here's how:
Step 1: Rearrange the equation:
2cos^2(theta) + 5sin(theta) - 4 = 0
Step 2: Since cos^2(theta) + sin^2(theta) = 1, we can substitute cos^2(theta) with 1 - sin^2(theta):
2(1 - sin^2(theta)) + 5sin(theta) - 4 = 0
Step 3: Distribute and rearrange the equation:
2 - 2sin^2(theta) + 5sin(theta) - 4 = 0
-2sin^2(theta) + 5sin(theta) - 2 = 0
Step 4: Factor the quadratic equation:
(-2sin(theta) + 1)(sin(theta) - 2) = 0
Step 5: Set each factor equal to zero and solve for theta:
-2sin(theta) + 1 = 0 --> -2sin(theta) = -1 --> sin(theta) = 1/2 --> using inverse sin function, we get theta = π/6 or 30 degrees
sin(theta) - 2 = 0 --> sin(theta) = 2 (Not possible, as the sine value is bounded between -1 and 1)
Therefore, the solution to the equation 2cos^2(theta) + 5sin(theta) - 4 = 0 is theta = π/6 or 30 degrees.
2(1-sin^2) + 5sin - 4 = 0
2sin^2 - 5sin + 2 = 0
(2sin-1)(sin-2) = 0
sin = 1/2 or 2
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