Find m to cos²x-(m²-3)sinx+2m²-3=0 have root

would ensure that at least one real root exists for this equation when solving for .

Step-by-step explanation:

Apply the Pythagorean identity to replace the cosine this equation with sine:

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Multiply both sides by to obtain:

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If , then this equation would become a quadratic equation about :

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However, for all real .

Hence, the value of must be between and (inclusive) for the original equation to have a real root when solving for .

Determinant of this quadratic equation about :

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Hence, when solving for , the roots of in terms of would be:

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Since , it is necessary that for the original solution to have a real root when solved for .

The first solution, , does not meet the requirements. On the other hand, simplifying , gives:

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In other words,  solving for would give a real root between if and only if .

On the other hand, given that for the in the original equation, solving that equation for would give a real root if and only if .

Therefore, the original equation with as the unknown has a real root if and only if .

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step-by-step explanation:

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5