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How do you solve 6cos2 theta plus 5cos theta minus 4 equal to 0?
Wiki User
∙ 11y ago
Assume the first term is cos-squared(theta) rather than cos(2*theta).
6cos2(t) + 5cos(t) - 4 = 0
[6cos2(t) + 8cos(t) - 3 cos(t) - 4] = 0
[3cos(t) + 4]*[2cos(t) - 1] = 0
which gives
3cos(t) = -4 so that cos(t) = -4/3
or
2cos(t) = 1 so that cos(t) = 1/2
The first of these is clearly not a possible solution, whereas the second can hve one or more solutions, depending on the domain - which is not specified in the question.
If the initial assumption is incorrect and the first term WAS 6cos(2*theta) then using the formula for double angles:
cos(2(t) = cos2(t) - sin2(t) = cos2(t) - [1 - cos2(t)] = 2cos2(t) - 1
and so, the equation becomes
6*[2cos2(t) - 1] + 5cos(t) - 4 = 0
or 12cos2(t) + 5cos(t) - 10 = 0
Solve this quadratic equation for cos(t) and then find the values of t (or theta) within the domain specified.
Wiki User
∙ 11y ago
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Since there is no equation, there is nothing that can be solved.
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Zero. Anything minus itself is zero.
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It is 1.
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It is a mathematical expression.
Can you solve sin 2 theta equals 0?
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There is no easy simplification.
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Since there is no equation, there is nothing that can be solved.
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cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
What is Cos Theta minus Cos Theta?
Zero. Anything minus itself is zero.
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It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
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- cos theta
What is cos 90 minus theta?
cosine (90- theta) = sine (theta)
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To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)
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-2(cot2theta)
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Tan^2
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If r-squared = theta then r = ±sqrt(theta)
What is sine squared theta subtract six cosine squared theta divided by one minus seven cosine squared theta?
It is 1.
