Solve the following trigonometric equation over the interval of [0,2π). Show your work, thank you!
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Step-by-step explanation:

we would like to solve the following equation:

to do so we need to put the equation in term of a function of a angle therefore rewrite tanθ:

simplify multiplication:

cross multiplication:

isolate -3cosθ to left hand side and change its sign:

well we can rewrite sin²θ by applying Pythagorean theorem and that yields:

distribute:

now notice that our equation ended up with a pattern and that is Quadratic thus by letting cosθ be x to transform the equation:

rearrange it to standard form:

divide both sides by -1:

well we can factor the quadratic to do so rewrite -3x as -4x+x:

factor out 2x:

factor out 1:

group:

by Zero Product property we obtain:

cancel 1 from the first equation and add 2 to the e equation:

substitute back:

since cosθ is only defined for the interval [1,-1] the second equation is false for any value of θ but we can still continue the first equation

divide both sides by 2:

by unit circle we acquire:

add period of 2nπ:

for the interval [0,2π) θ is only defined when n is 0 Thus:

simplify:

hence,

θ is equal to 150° and 240°

step-by-step explanation:

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry