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Solve the equation; 4 cos^2 (x) + 7 sin (x) – 7 = 0, giving all answers between 0° and 360°.

4 cos2 (x) + 7sin (x) – 7 = 0      (1)

Using the identity sin2 (x) + cos2 (x) = 1,

rearrange to give; cos2 (x) = 1 - sin2 (x)     (2)

Sub (2) into (1)...

Giving;  4 ( 1 - sin2 (x) ) + 7sin (x) - 7 = 0

Expand brackets; 4 - 4sin2 (x) + 7sin (x) - 7 = 0

Simplify; 7sin (x) - 4sin2 (x) - 3 = 0

Let sin(x) = A and treat as a quadratic;

7A - 4A2 - 3 = 0

- (4A - 3)(A - 1) = 0

4A - 3 = 0 or A - 1 = 0

4A = 3  or A = 1

A = 3/4

However, we already have defined A = sin (x), therefore we must not stop here, but set A = sin (x) and solve for x as follows:

sin (x) = 3/4 or sin (x) = 1

x = sin-1 (3/4) or x = sin-1 (1)

x = 48.6° and (180 - 48.6)° [1] or 90°

x = 48.6° and 131° or 90.0° (to 3 significant figures)

[1]: Note that we take 180° - angle, as for x in sin (x) between 0° and 180° is symmetrical around 90°​. If x was between 180° and 360°, we'd do (360 - x) if 180 < x < 270 and (180 + x) if 270 < x < 360.

Connor S. GCSE Maths tutor, A Level Maths tutor, GCSE Physics tutor, ...

9 months ago

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