Answer:
(-1.570796, 0.666239) ∪ (1.570796, 2.475353)
Explanation:
A graphing calculator can give you the numerical values that limit the intervals of the solution. The attachment shows this solution.
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algebraic solution
We can multiply by cos(x) and solve the resulting quadratic inequalities. Two cases arise:
tan(x) -cos(x) < 0 . . . . . . subtract cos(x)
sin(x) -cos(x)² < 0 . . . . . . . for cos(x) > 0
sin(x) -cos(x)² > 0 . . . . . . . for cos(x) < 0 (inequality is reversed)
Replacing cos(x)² with (1 -sin(x)²) gives quadratic equations in sin(x). (We're using <> to mean "could be greater than; could be less than, depending on cos(x)".)
sin(x) -(1 -sin(x)²) <> 0
sin(x)² +sin(x) -1 <> 0 . . . . put in standard form
(sin(x)² +sin(x) +1/4) -1 -1/4 <> 0 . . . . complete the square
(sin(x) +1/2)² -5/4 <> 0 . . . . . write as a square
((sin(x) +1/2)-√(5/4))·((sin(x) +1/2) +√(5/4)) <> 0 . . . . factor
Note that the second factor is always positive, so it has no impact on the intervals of solution.
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The zeros of the first factor are located at ...
sin(x) = 1/2(-1 +√5)
x ≈ 0.666239 radians, and π - 0.666239 radians ≈ 2.475353 radians
The sine function will be positive between these angle values.
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However, the intervals of solution to the inequality depend on the cosine function. Effectively, the solution space is where (sin(x) +(1-√5)/2) and cos(x) have opposite signs.
cosine positive
The cosine function is positive for -π/2 < x < π/2, so one of the intervals of solution will be ...
-π/2 < x < arcsin((√5 -1)/2), approximately (-1.570796, 0.666239)
cosine negative
The cosine function is negative for -π < x < -π/2 and for π/2 < x < π. The other interval of solution will be ...
π/2 < x < (π -arcsin((√5 -1)/2)), approximately (1.570796, 2.475353)
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Additional comment
In the above, we have used α=arcsin( ) to mean the values of angle in the principal branch: -π/2 ≤ α ≤ π/2. That is why we have to subtract it from π to find the other end of the interval of interest.