Question

1) 2cosx(cosx-sinx)^2 2) 3sin^2 2x+7cox2x-3=0 3) sin^2 x-4sinx>0

Answer 1

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Answer 2

Let's solve each of the given equations one by one:

1. 2cos(x)(cos(x) - sin(x))^2:

To solve this equation, we'll simplify it step by step:

Expand the squared term: 2cos(x)(cos^2(x) - 2cos(x)sin(x) + sin^2(x))

Simplify using the identity cos^2(x) + sin^2(x) = 1: 2cos(x)(1 - 2cos(x)sin(x))

Now we have: 2cos(x) - 4cos^2(x)sin(x)

2. 3sin^2(2x) + 7cos(2x) - 3 = 0:

This is a trigonometric equation involving both sine and cosine functions. We'll use the double-angle identities to simplify it:

Using the double-angle identities, we have: 3(1 - cos^2(2x)) + 7cos(2x) - 3 = 0

Expand the square: 3 - 3cos^2(2x) + 7cos(2x) - 3 = 0

Combine like terms: -3cos^2(2x) + 7cos(2x) = 0

3. sin^2(x) - 4sin(x) > 0:

We'll solve this inequality step by step:

Factor out sin(x): sin(x)(sin(x) - 4) > 0

Now we have two factors: sin(x) and (sin(x) - 4). To determine the sign of the inequality, we'll analyze the sign of each factor separately.

For sin(x) > 0, the solution lies in the intervals where sin(x) is positive: (0, π) and (2π, 3π), etc.

For sin(x) - 4 > 0, the solution lies in the interval where sin(x) is greater than 4, which is not possible since the range of sin(x) is [-1, 1].

Therefore, the solution is sin(x) > 0, which corresponds to the intervals (0, π) and (2π, 3π), etc.

Please note that the given equations and inequalities may have more specific solutions depending on the domain or range specified. The above solutions provide a general approach to solving them.