Math Methods problem (Trig question)

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To solve the equations 2cosX + tanX = secX and 2sinX + cotX = cscX, one can use trigonometric identities and algebraic manipulation. For the first equation, rewriting secX as 1/cosX and applying the identity cos²X + sin²X = 1 helps simplify the problem. This leads to a quadratic equation in terms of sinX, which can be factored to find solutions. The valid solution for sinX from the first equation is -1/2, while sinX = 1 is not applicable within the specified domain. Understanding these identities and transformations is crucial for solving trigonometric equations effectively.
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I have 2 problems. First, solve 2cosX + tanX = secX where X is greater than or equal to zero and less than 2 pi. I understand that the secant is basically the inverse of the cosine, (hyp/adj), but I have no idea how to solve for this. I also don't understand how to solve an equation with 2 different trig functions in it (sin and cos)
Please help!

Also, it says solve 2sinX + cotX = cscX where X is defined by the same parameters as mentioned above. I have the same problem. Help! Thanks!
 
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A few hints:
1. Multiply your equation with cos(x).
2. Use a well-known identity to exress cos^(2)(x) in terms of sin(x)
3. Solve for sin(x)
 
Basically arilno's ideas:

2cos(x) + tan(x) = 1/cos (x)

2cos^2 (x) + cos(x) (sin(x) / cos(x)) = 1

The well known identities: cos^2(x) + sin^2 (x) = 1, tan(x) = sin(x) / cos(x).

2 (1 - sin^2 (x)) + sin(x) = 1

2 - 2sin^2 (x) + sin(x) = 1

2sin^2 (x) - sin(x) - 1 = 0

This is a quadratic equation.

(2 sin(x) + 1)*(sin(x) - 1) = 0

sin(x) = -1/2 or sin(x) = 1
 
sinx =1 is not applicable coz it is not in the domain
 
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