Question
14. Solve the equation 2tanx^circ +5=-4 ,for 0leqslant xleqslant 360
Answer
4
(430 Votes)
Wynona
Master · Tutor for 5 years
Answer
Follow calculations reflected in the explanaition, the computation implies cut point locations fallen, suggesting potential angles
x = 360^{\circ} \pm x' \) (modulo 180^\circ). There might be multiple possible solutions or cases depending on coterminal angles and differences falling within the stipulated inclusive range of
.Trying these out:Giving approach that trying out \(\tan^{-1} (\frac{-9}{2})\) we might find
approximately equaling the overall to -77.471192^{\circ}. Following, flipping the sign could contribute the results of positive values. Working the other values on the othe hand out we might find say are (subtracting achieved values, \(x':x = 180^{\circ} - (-77.471192) =257.471192 \) and the next possible one says\( x = 360^{\circ} - (-77.471192) = 437.471192 \). But, do realise that the value x doesn't fall within the inclusive range reasoned out initially. After reductive actions trying to revoke modulo 180^\circ from 437.471 the result implies possible
. Following these claims as final answer denotes through thus reflecting clearly explanative solving paradigm.
Explanation
## Step 1: Simplify the EquationFirst unnest the equation
. To do this, we simplify it by moving 5 on the other side. For this, subtract 5 from both sides of the equation to get this equation:
.### The formula we have now:
.## Step 2: Solve for xEither divide both sides of the equation by 2 or consider it as a seperation of '2', to leave out 'tan x' on the left shift of the equation. This step can be formed as
or
.## Step 3: After taking
We need to identify where our value for
lies within the specified range. Inverse tangent is equal to our shifted variable from the last step (i.e.
). Do note, calculators often provide a result within a restricted range usually within
, We need to take this by finding out which quadrant our actual solution falls in based on the knowledge that tangent is negative in the second and fourth quadrants (since sine and cosine are positive only in the first and second quadrants).