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say (2x-π/6)=qtan²q = sec²q-1so sec²q-1= 11 - sec q sec²q + sec q -12 = 0(sec q -3) (sec q + 4) = 0sec q = 3 or -4because 0<=x<= πso -π/6 <= q <= 11...

e^ln(2x-3) = e¹
2x-3 = e
x = (e+3)/2

Assuming knowledge of solving regular quadratic equations.Treat this as a regular quadratic equation, except it is in terms of sin(x) instead of just x as you might be used to.So using the quadratic equat...

To answer this question you must use implicit differentiation due to there being both x and y terms. Consequently you must differentiate each term individually as you would usually (by multiplying by the ...

When dealing with a linear and quadratic simultaneous equation, it is best to use subsitution to approach it.I take the linear equation and put it in terms of one variable. In this case, to avoid fraction...